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ELEMENTS OF
ENGINEERING
THERMODYNAMICS
BY
JAMES A. MOYER
Director of the Massachusetts Department of University Extension
formerly Professor of Mechanical Engineering in the
Pennsylvania State College
JAMES P. CALDERWOOD
Professor of Mechanical Engineering in the Kansas State
Agricultural College
ANDREY A. POTTER
Dean of Engineering at Purdue University, formerly
Dean of Engineering at the Kansas State
Agricultural College
NEW YORK
JOHN WILEY & SONS, Inc.
London: CHAPMAN & HALL, Limited
1920
< 3^
Copyright, 1920, by
JAMES A. MOYER, JAMES P. CALDERWOOD
AND
ANDREY A. POTTER
£o/t$te
Manufactured in the U.S.A.
CCT I5IS20
©CI.A576885
PREFACE
This treatise is an extension of a briefer work entitled " Engi
neering Thermodynamics" by James A. Moyer and James P.
Calderwood. The additions and changes are made to suit the
needs of those who had successful experience in using the original
publication and found it desirable to add supplementary material
to make it sufficiently inclusive for special institutional require
ments. A great deal of the new material is supplied by A. A.
Potter.
This book is intended to bring out the fundamental principles
of Engineering Thermodynamics, and is particularly intended
for use in technical colleges where it is possible to give special
courses on the subjects of steam turbines, internal combustion
engines, refrigeration and other applications of thermody
namics.
The new material includes the theory of the hot air engine
and internal combustion engine cycles. The appendix includes
logarithmic tables, the properties of gases, the properties of
steam, ammonia, sulphur dioxide and carbon dioxide.
Every engineering student should become familiar with stand
ard works on the subject of thermodynamics. This book should
consequently be supplemented by references to standard works on
this subject. A representative list of such reference books is
given in a table of the appendix.
The authors are particularly indebted to the following pro
fessors and instructors in Mechanical Engineering Departments
who gave valuable suggestions and criticisms: Edwin A. Fessen
den, Pennsylvania State College; J. E. Emswiler, University of
Michigan; G. L. Christensen, Michigan College of Mines; Roy
in
IV
PREFACE
B. Fehr, formerly of Pennsylvania State College; H. L. Seward,
Yale University; J. R. Wharton, University of Missouri; and
J. J. Wilmore, Alabama Polytechnic Institute.
J. A. MOYER,
J. P. CALDERWOOD,
A. A. POTTER.
CONTENTS
Pages
Chapter I. — Thermodynamic Principles and Definitions 113
Definition, Scope and Object of Thermodynamics; Heat; Temperature;
Thermometers; Absolute Zero; Units of Heat; Specific Heat; Pressure;
Volume; Work; Power; Indicated Horsepower; Brake Horsepower;
Mechanical Equivalent of Heat; First Law of Thermodynamics; Second
Law of Thermodynamics; Effects of Heat Application; The Heat Engine;
Thermal Efficiency of a Heat Engine; Problems.
Chapter II. — Properties of Perfect Gases 1425
Perfect Gas Defined; Relation between Pressure, Volume, and Temper
ature of a Perfect Gas; Boyle's Law; Charles' Law; Combination of
Boyle's and Charles' Laws; The Law of Perfect Gases; Heat and its
Effect upon a Gas; External Work; Internal Energy; Joule's Law;
Relation of Specific Heats and the Gas Constant; Ratio of the Two
Specific Heats; Values of the Specific Heats; Problems.
Chapter III. — Expansion and Compression of Gases 2639
Expansion at Constant Pressure; Expansion at Constant Volume;
Expansion at Constant Temperature, or Isothermal; Adiabatic Expan
sion and Compression; Change of Internal Energy during Adiabatic
Processes; Relation between Volume, Pressure and Temperature in
Adiabatic Expansion of a Perfect Gas; Problems.
Chapter IV. — Cycles of Heat Engines using Gas 4061
(1) The Carnot Cycle; Reversible Cycles.
(2) Hot Air Engine Cycle. The Stirling Engine; The Ericsson Engine.
(3) Internal Combustion Engine Cycles. The Lenoir Engine; The Otto
Cycle; The Brayton Cycle; the Diesel Cycle.
Problems.
Chapter V. — Properties of Vapors 6290
Saturated and Superheated Vapors; Theory of Vaporization; Vapor
Tables; Relation between Temperature, Pressure and Volume of Satu
rated Steam; Heat in the Liquid (Water); Latent Heat of Evaporation;
External Work of Evaporation; Total Heat of Steam; Internal Energy
of Evaporation and of Steam; Steam formed at Constant Volume; Wet
Steam; Superheated Steam; Drying of Steam by Throttling or Wire
VI CONTENTS
i • ^ • , , Pages
drawing; Determination of the Moisture in Steam; Throttling or
Superheating Calorimeter; Barrus Throttling Calorimeter; Separating
Calorimeters; Condensing or Barrel Calorimeter; Equivalent Evapo
ration and Factor of Evaporation; Vapors as Refrigerating Media;
Problems.
Chapter VI. — Entropy 91103
Entropy Changes during Constant Pressure Expansions of Gases;
Entropy Changes of Gases at Constant Volume; Entropy Changes
during Isothermal Processes of a Gas; Entropy Changes during Reversible
Adiabatic Processes of Gases; Entropy Changes during a Carnot Cycle;
TemperatureEntropy Diagrams for Steam; Calculation of Entropy for
Steam; Total Entropy of Steam; The Mollier Chart; Temperature
Entropy Diagram for the Steam Power Plant; Problems.
Chapter VII. — Expansion and Compression of Vapors 104116
Expansion of Wet Steam at Constant Volume; Expansion of Super
heated Steam at Constant Volume; Expansion at Constant Pressure;
Isothermal Lines of Steam; Adiabatic Lines for Steam; Quality of
Steam during Adiabatic Expansion; Graphical Determination of Quality
of Steam by Throttling Calorimeter and Total HeatEntropy Diagram;
Poly tropic Expansion (n = 1); Problems.
Chapter VIII. — Cycles of Heat Engines using Vapors 11 7149
Carnot Cycle; Rankine Cycle; Practical or Actual Steam Engine Cycle;
Efficiency of an Engine Using Steam Without Expansion; Adiabatic
Expansion and Available Energy; Available Energy of Wet Steam;
Available Energy of Superheated Steam; Application of Temperature
Entropy Diagram to Analysis of Steam Engine; Combined Indicator
Card of Compound Engine; Hirn's Analysis; Clayton's Analysis; Prob
lems.
Chapter IX. — Flow of Fluids 150177
Flow Through a Nozzle or Orifice; Weight per Cubic Foot; Maximum
Discharge; Shape of Nozzle; Flow of Air Through an Orifice; Receiver
Method of Measuring Air; Flow of Vapors; Velocity of Flow as Affected
by Radiation; Friction Loss in a Nozzle; Impulse Nozzles; Turbine
Losses; Reaction Nozzles; Coefficient of Flow; Injectors; Weight of
Feed Water Supplied by an Injector per Pound of Steam; Thermal Effi
ciency of an Injector; Mechanical Efficiency of an Injector; Orifice
Measurements of the Flow of Steam; Flow of Steam Through Nozzles;
Flow of Steam when the Final Pressure is more than 0.58 of the Initial
Pressure; Length of Nozzles; Efficiency of Nozzles; Under and Over
Expansion; Nonexpanding Nozzles; Materials for Nozzles; Problems.
CONTENTS VU
Page
Chapter X. — Applications of Thermodynamics to Compressed Atr
and Refrigerating Machinery 178196
Compressed Air; Air Compressors; Work of Compression; Effect of
Clearance upon Volumetric Efficiency; Two Stage Compression; Re
frigerating Machines or Heat Pumps; Unit of Refrigeration; Systems
of Mechanical Refrigeration; The Air System of Refrigeration; The
Vapor Compression System of Refrigeration; The Vapor Absorption
System of Refrigeration; Coefficient of Performance of Refrigerating
Machines; Problems.
SYMBOLS
A = area in square feet, also used to represent the reciprocal of the mechanical
equivalent of heat, yfg.
B.t.u. = British thermal units (= 778 ft.lbs.).
Cp = specific heat at constant pressure in B.t.u. per pound per degree.
Cv = specific heat at constant volume in B.t.u. per pound per degree.
C = a general constant in equations of perfect gases.
E = external work in B.t.u. per pound; also sometimes used to express effi
ciency, usually as a decimal.
E a = available energy in B.t.u. per pound.
F = force in pounds.
H = heat per pound in B.t.u.*
Hsup. = total heat of superheated steam, B.t.u. per pound.
In = total internal energy of steam (above 32 F.) in B.t.u. per pound, some
times designated by E and i".
I L = internal energy of evaporation of steam in B.t.u. per pound, sometimes
designated by p.
J = mechanical equivalent of heat = 778 (use becoming obsolete).
K = specific heat in footpound units.
L = latent heat of evaporation in B.t.u. per pound.
P = pressure in general or pressure in pounds per square foot.
Q = quantity of heat in B.t.u.
R = thermodynamic constant for gases; for air it is 53.3 (in footpound units
per pound.)
T = absolute temperature, in Fahr. degrees = 460 + t.
V = volume in cubic feet, also specific volume and velocity in feet per second.
W = work done in footpounds.
a = area in square inches.
c = constant of integration.
d = distance in feet.
e = subscript to represent base of natural logarithms.
g = acceleration due to gravity = 32.2 feet per second per second.
h = heat of the liquid per pound in B.t.u. (above 32 ° F.) also designated
by q and i' .
i' = heat of liquid in B.t.u. per pound (above 32 F.).
i" = total internal energy of steam.
k = a constant,
log = logarithm to base 10.
loge = logarithm to natural base e (Naperian).
* In steam tables it is total heat above 32° F.
ix
SYMBOLS
n = general exponent for V (volume) in equations of perfect gases, also some
times used for entropy of the liquid in B.t.u. per degree of absolute tem
perature.
p = pressure in pounds per square inch.
q = sometimes used for heat of the liquid in B.t.u. per pound (above 3 2° F.).
r = ratio of expansion also sometimes used for latent heat of evaporation in
B.t.u. per pound.
s' = entropy of the liquid.
s" = total entropy of vapors.
t = temperature in ordinary Fahr. degrees.
u = difference between the specific volume of a vapor and that of the liquid
from which it is formed.
v = specific volume, in cubic feet per pound (in some steam tables) .
w = weight per cubic foot = density, also used to designate weight of a sub
stance in pounds.
x = quality of steam expressed as a decimal.
r<
y = ratio of specific heats — .
Cp
4> = total entropy ^f, sometimes designated by S".
6 — entropy of the liquid in B.t.u. per pound per degree of absolute tem
perature,
p = internal energy of evaporation, B.t.u. per pound.
<r = volume of water in a cubic foot of saturated steam.
" — inches.
ELEMENTS OF
Engineering Thermodynamics
CHAPTER I
THERMODYNAMIC PRINCIPLES AND DEFINITIONS
Thermodynamics. Thermodynamics deals with the relation
between heat and mechanical work. The object of the study of
thermodynamics is to consider factors which influence the
efficiency of heat power machinery.
Thermodynamics makes it possible to predict the perform
ance of steam engines and steam turbines when operating under
conditions of increased pressure, of higher vacuums and expan
sions, of higher superheats, of jacketing cylinders, of using the
working medium in several cylinders one after the other, — that
is, compounding instead of expanding the steam only in a single
cylinder, — of inserting receivers between the cylinders and of
reheating the working medium as it passes from one cylinder to
the next.
By means of calculations based upon the study of thermo
dynamics it is possible to determine the effect of increasing the
compression pressures of a gas engine mixture before it is
ignited, the result of incomplete cooling upon the efficiency of
an air compressor, and similar problems.
Another important service which the study of thermodynamics
renders is that of showing what maximum efficiency is attain
able for any heat engine operating under a given set of conditions.
It often happens that tests indicate an efficiency very much bet
ter than is usually obtained with any of the present types of
engines. In such cases thermodynamic calculations will show
conclusively whether the results secured are possible. The
ability to interpret correctly the results of experiments performed
on all kinds of heat engines requires a knowledge of the basic
principles of thermodynamics.
2 THERMODYNAMIC PRINCIPLES AND DEFINITIONS
Heat. Heat is a form of energy and not a material substance.
The heat of a body depends on the vibratory motion of the small
particles or molecules of which the body is built up, the greater
the velocity and amplitude of the vibration of these molecules,
the higher is the temperature of the body.
Heat can be transferred by conduction, radiation and con
vection.
The transfer of heat between the different particles of the
same body or between several bodies is called conduction.
In a heated body the particles are in violent agitation and as
a result waves are formed and are emitted or are radiated
through space to other bodies.
The transfer of heat by the motion of the heated particles is
called convection. This phenomenon is exhibited in liquids and
gases. Thus, when a hot body heats the air in contact with it,
the currents of air which are produced by the process of con
vection ascend and are replaced by cooler air.
Temperature. Temperature is the indication of the sensible
heat of a substance and can be measured by a thermometer. The
temperature does not measure the quantity of heat energy in the
substance, but indicates only the relative heat intensity, which
can be revealed by the senses of the observer.
Thermometers. There are three thermometric scales:
The Centigrade or Celsius degree is T fo of the temperature in
terval between the melting point of ice and the boiling point of
water at atmospheric pressure, these two fixed points being de
noted o° C. and ioo° C. respectively.
The Fahrenheit degree is T §o of the temperature interval
between these two fixed points, the melting point of ice being
taken at 3 2° F. and the boiling point of water at 212 F.
The Reaumur scale has the melting point of ice at o° R. and the
boiling point of water at 8o° R.
The thermometric scales generally used are the Centigrade
and Fahrenheit, the relations between these scales being:
Degrees C. =  [degrees F. — 32]. (1)
Degrees F. = f degrees C. + 32. (2)
UNITS OF HEAT 3
Mercury thermometers, as ordinarily constructed, have the
space above the mercury under a vacuum. Such thermometers
cannot be used for the measurement of temperatures exceeding
500 F., as the vacuum reduces the boiling point of mercury.
The range of mercury thermometers can be increased to about
900 F. by filling the space above the mercury with some inert
gas like nitrogen.
For the measurement of very high temperatures thermo
electric pyrometers are best suited.
Absolute Zero. In the graduation of liquid thermometers, the
vaporization of the liquid at high temperatures and its freezing
at low temperatures limits the thermometric range. The funda
mental scale of temperature measurement is based on Thom
son's absolute thermometric scale, which is independent of the
nature of any thermometric substance. The zero of the abso
lute scale or the absolute zero is taken as a point which marks the
absence of heat energy or of molecular vibrations of a body.
The absolute zero is 459. 5 (practically 460 ) below the zero
on the Fahrenheit scale and 273.0 below the zero on the Centi
grade scale.
Calling the absolute temperature T and the temperature as
measured by a thermometer /,
On the Fahrenheit scale T = t + 460. (3)
On the Centigrade scale T = / + 273. (4)
Units of Heat. Heat is measured in heat units. A heat unit
is the amount of heat required to raise the temperature of one
unit weight of water one degree. In the English system of
measures the heat unit is the British thermal unit (B.t.u.), which
is defined as the amount of heat required to raise the tempera
ture of one pound of water one degree on the Fahrenheit scale.
The heat unit in the metric system is the calorie * which is de
fined as the heat required to raise the temperature of one kilo
gram of water one degree on the Centigrade scale.
* Since one kilogram = 2.204 pounds and one degree C. = ° F., 1 calorie
= f X 2.204 = 3968 B.t.u.
4 THERMODYNAMIC PRINCIPLES AND DEFINITIONS
To correctly define the British thermal unit it is necessary
to state at what temperature the rise of one degree on the Fahren
heit scale is to occur, because the specific heat of water is slightly
variable. This heat unit (B.t.u.) is sometimes denned as the
amount of heat required to raise the temperature of water one
degree Fahrenheit at the condition of maximum density of water,
that is, between 39 and 40 degrees Fahrenheit. Other definitions
are based on the amount of heat required to raise the temperature
of water one degree Fahrenheit between 60 to 61 degrees Fahren
heit. Still another definition, which is generally considered to be
the most accurate, defines a British thermal unit as one one
hundredandeightieth ( T §~o) of the amount of heat required to
raise the temperature of water from 32 to 212 degrees Fahrenheit.
In other words, according to this last definition, the British ther
mal unit is the average value of the amount of heat required to
raise the temperature of one pound of water one Fahrenheit
degree between the conditions of freezing and boiling at atmos
pheric pressure.
Specific Heat. As the addition of the same quantity of heat
will not, as a rule, produce the same temperature changes in equal
weights of different substances, it is evident that the amount of
heat in any substance will depend on the capacity of the substance
for heat. For this reason it is necessary to allow for the relative
heat capacity or the specific heat (C) of a substance. Specific
heat can be defined as the ratio of the heat added to the tempera
ture change produced in a unit weight of a substance. It can
also be defined as the resistance which a substance offers to a
change in its temperature, the resistance of water being taken as
unity. In the English system the specific heat is the number of
British thermal units (B.t.u.) required to raise the temperature
of a pound of the substance one degree Fahrenheit.
Thus if Q is the quantity of heat added to one pound of a sub
stance, the temperature change is,
h\=% (5)
or Q = C (fe  h).
SPECIFIC HEAT 5
If the specific heat is a variable,
Cdt. (6)
The following problem illustrates the application of equation
(6):
The specific heat of a substance is expressed by the equation,
C = 0.241 1 2 + 0.000009 t.
What amount of heat is required to raise the temperature of
one pound of the substance from o° to ioo° F.?
Solution. Since the specific heat is variable,
h
Cdt.
substituting the value of C and integrating,
(0.241 1 2 + 0.000009 1) &
h
n ioo° , nn 1000
= O.24I 1 2 [/J QO + O.OOOOO9 
L 2 Jo°
= O.24I 1 2 (lOO) + O.OOOOO9 (5000)
= 24.112 + 0.045 = 24.157 B.t.u.
The specific heat of gases and vapors changes considerably in
value according to the conditions under which the heat is applied.
If the heat is applied to a gas or a vapor held in a closed vessel,
with no change in volume, no work is performed, and, therefore,
all the heat added is used to increase the temperature. This is
the condition in a boiler when no steam is being drawn off. In
this case the symbol C v represents the specific heat during heat
application at constant volume. If, on the other hand, the heat
ing is done while the pressure is kept constant and the volume is
allowed to change permitting expansion and the performance of
work, the symbol C v is used and represents the specific heat dur
ing heat application at constant pressure.
6 THERMODYNAMIC PRINCIPLES AND DEFINITIONS
When the problem deals with w pounds of a substance instead
of a unit weight, equation (6) becomes,
Q = wf h Cdt. (7)
Pressure. Force per unit of area is called pressure. Thus,
the pressure exerted by a gas or vapor is expressed in the
English system in pounds per square inch, pounds per square
foot, inches of mercury or atmospheres. In the metric system,
pressure is expressed in kilograms per square centimeter or
millimeters of mercury.
Gages always read pressures above atmospheric pressure or
above vacuum. The absolute pressure is the sum of the gage
and atmospheric pressures. Thus, if a gage reads 75 pounds
pressure (per square inch), and the barometer is 29.65 inches of
mercury, the atmospheric or barometric pressure is, in pounds per
square inch,
29.65 X 0.491 = 14.56.
(0.491 is the weight of a cubic inch of mercury at 70 F.)
The absolute pressure is:
75 + 14.56 = 89.56 pounds per square inch.
In thermodynamic equations the unit of pressure is usually
expressed in pounds per square foot.
Volume. By specific volume is meant the amount of space
occupied by a unit weight of a substance, expressed in cubic feet
or in cubic meters. Thus, the volume of one pound of steam at
atmospheric pressure is its specific volume and is equal to 26.79
cubic feet.
Work. Work is done by a force during a given displacement
and is independent of the time. The footpound is the unit of
work in the English system. Thus, when a body weighing one
pound is raised through a distance of one foot, the resulting work
is a footpound. Similarly, the product of the pressure in pounds
POWER
per square foot and the volume in cubic feet is equal to work in
footpounds.
Space
Fig. i. — Work Diagram.
Work being the product of two dimensions, it may be repre
sented graphically by the area of a closed figure, the coordinates
of which are force and distance, or pressure and volume. Thus,
A (Fig. i ) represents the position of a body between F (force) and
5 (space) coordinates, being the origin or starting point. The
distance A D f rpm the OS line represents the force against which
it is acting and AA', measured from the OF line, is the distance it
has moved from the starting point. If the body moves from A
to B along the path AB, the area A BCD represents the work
done.
If the equation of the path is known, the work is
W
£"
SB
FdS.
(8)
Power. Power is the rate of doing work, or the work done
divided by the time required to do it. In the English system, the
unit of power is the horse power. It is the power required to
raise 550 pounds through a vertical distance of one foot in one
8 THERMODYNAMIC PRINCIPLES AND DEFINITIONS
second, or 33,000 pounds one foot in one minute. To obtain the
horse power, the work in footpounds per minute must be obtained
and the result divided by 33,000. In the metric system, the unit
of power is the watt. One horse power is equal to 746 watts.
The French horse power (cheval) is 542.5 footpounds per second.
Indicated Horse Power. The term indicated horse power is
applied to the rate of doing work by a gas or a vapor in the cylin
der of an engine. It is obtained by means of an engine indicator.
If P represents the mean effective pressure (average unbal
anced pressure) in pounds per square inch, as shown by the
indicator card, and A the effective area of the piston in square
inches, the total pressure exerted on the piston is PA. If the
piston has a stroke of L feet, the work per stroke is PAL and the
work per minute is PA L N, where N represents the number of
revolutions per minute. The indicated horse power is
T , Work per minute PA LN
I.h.p. = £ = . (9)
33,000 33>°°°
Brake Horse Power represents the actual power which an
engine can deliver for the purposes of work. The difference
between indicated and brake horse power of an engine represents
the horse power lost in friction. The brake horse power can be
measured by some form of friction brake, which absorbs the power
measured, and is called an absorption dynamometer, or by a
transmission dynamometer. In either type of dynamometer,
if F is the effective pull in pounds, L the lever arm, in feet,
through which the weight is exerted, and N the number of revo
lutions of the shaft per minute, the brake horse power is
,, 2TFLN , ,
B.h.p. = (10)
33>°°° •
Mechanical Equivalent of Heat. There is a definite quantita
tive relation between work expended and heat produced. This
relation between heat and work is called the mechanical equiva
lent of heat and is designated by /. In the English system,
J = 778 footpounds
or 1 B.t.u. =778 footpounds.
EFFECTS OF HEAT APPLICATION 9
In the metric system,
/ = 427 kilogrammeters
or 1 calorie = 427 kilogrammeters.
The reciprocal of J, or the heat equivalent of work, is designated
by A. where A = —  in the English system or in the metric
778 6 427
system.
First Law of Thermodynamics. The statement of the definite
relation between heat and mechanical work constitutes what is
known as the first law of thermodynamics. It is usually
expressed:
"Heat and mechanical energy are mutually convertible and
heat requires for its production and produces by its disappear
ance mechanical work in the ratio of the mechanical equivalent
of heat." In other words, this law is a statement of the conserva
tion of energy as regards the equivalence of mechanical work and
heat.
Second Law of Thermodynamics. "In order to transform the
heat of a body into work, heat must pass from that body into
another at a lower temperature." Thus, there must be a dif
ference of level in the transformation of heat energy into work
and heat cannot be transformed from one body to another at a
higher temperature, unless work is expended in order to produce
such a transfer of heat.
This law states as regards heat engines the limits to their pos
sible performance, which would be otherwise unlimited, if only
the "first law" of thermodynamics is considered. It means, also,
that no heat engine converts or can convert into work all of the
heat supplied to it. A very large part of the heat supplied is
necessarily rejected by the engine in the form of unused heat.
Effects of Heat Application. If a quantity of heat Q is im
parted to a body, the following effects will be produced:
1. The temperature of the body will rise;
2. The volume of the body will increase;
3. The body will be capable of doing external work.
IO THERMODYNAMIC PRINCIPLES AND DEFINITIONS
Representing the above effects by S, V and W,
Q = S+V + W, (n)
in which S represents that quantity of heat, termed the sensible
heat, which was utilized in raising the temperature of the body; V
represents that quantity of heat which was absorbed by the body
in increasing its store of internal energy other than that associated
with the sensible heat; and W is the heat which was absorbed to
perform the external work in increasing the volume of the body
against the resistance offered by external substances. S + V
represents that portion of the heat that was chargeable to the
increase of internal work. This may be termed the intrinsic
energy increase and designated by /.
Thus, for the addition of an infinitesimal quantity of heat,
dQ = dI + dW
or Q =  ~dl  f *PdV. (12)
In equation (12), which represents the effect of heat application,
the intrinsic energy change / dl depends on the physical state
of the body, that is, whether it is a solid, a liquid, a vapor, or a
/■» Vt
P dV depends on the char
Vi
acter of the path in the work diagram.
The Heat Engine. The heat engine is a machine which con
verts the heat energy of solid, liquid or gaseous fuel into work.
This conversion depends on the variation in the pressure, volume
and temperature of a gas or a vapor and can be accomplished in
two ways:
First, by "external combustion," in which case the fuel is
burned outside of the engine cylinder; the heat developed by the
combustion of the fuel is conducted to the working substance or
heat medium through walls; this working substance does work
on a piston in the case of a reciprocating engine or on a vane or
THERMAL EFFICIENCY OF A HEAT ENGINE n
" blade " in a turbine. To this class belong steam engines of the
reciprocating, turbine or rotary types and also externalcombus
tion hot air engines. Thus, in the case of the steam engine, the
fuel, which may be coal, wood, petroleum or gas, is burned out
side of the boiler shell and the resulting heat is transmitted by
conduction through the metal of the shell to the working sub
stance, which is water. When enough heat has been added to the
water to produce a change in its physical state, water vapor or
steam is formed at the required pressure. This vapor, which
may be dry, wet or superheated, if allowed to act on the piston
of the engine, will do work.
Another method of converting heat into mechanical energy is
by burning the fuel rapidly or slowly inside of an engine cylinder
or in a communicating vessel, the products of combustion being
allowed to act directly on the piston of the engine. To this class
belong gas, petroleum and alcohol engines which are called
" internal combustion " engines.
Thermal Efficiency of a Heat Engine. By thermal efficiency
(E) is meant the ratio of the heat converted into work (AW) to
the heat supplied by the engine (Qi), or
77 AW
E  ^. (13)
Since only a part of the heat supplied to an engine can be con
verted into work, the above ratio is a fraction always less than
unity.
PROBLEMS
1. Give examples of the transfer of heat by conduction, radiation and
convection.
2. Show that the kilogram calories per kilogram X 1.8 give B.t.u. per
pound.
3. Prove that the product of a pressure in pounds per square foot and a
volume in cubic feet results in footpounds of work.
4. If the specific heat of a substance is 0.65, how many B.t.u. are re
quired to raise the temperature of 10 lbs. of the substance through io° F.?
5. The specific heat of a substance is expressed by the equation,
C = O.S — O.C2 t.
12 THERMODYNAMIC PRINCIPLES AND DEFINITIONS
What heat is required to raise the temperature of 5 lbs. of the substance
from io° to ioo° F.?
6. Convert — 40 C. into degrees Fahrenheit.
7. Change 350 F., 212 F. and 160 C. to absolute Fahrenheit tempera
tures.
8. Prove that the weight of 1 cu. in. of mercury is equal to 0.491 lb.
9. If the barometric reading is 29.2 ins., change 140 lbs. per sq. in. gage
and also 27 ins. vacuum into pounds per square inch absolute pressure.
10. Calculate the indicated horse power of a 12" X 13" steam engine
which operates at a speed of 265 r.p.m. The mean effective pressure of the
head end is 27.5 lbs. per sq. in. and of the crank end 27.8 lbs. per sq. in. The
diameter of the piston rod is 1 f inches.
11. Calculate the brake horse power developed by an engine as measured
by a Prony brake, the effective pull being 32 lbs. at 250 r.p.m. and the
lever arm 32 inches long.
12. One pound of fuel has a heating value of 14,500 B.t.u. How many
footpounds of work is it capable of producing, if all this heat is converted
into work?
13. An engine developed 15,560 footpounds of work. How much heat
in B.t.u. was theoretically required?
14. A heat engine receives 100,000 B.t.u. of heat in the form of fuel and
during the same period 30,000 B.t.u. are converted into work. What per
centage (thermal efficiency) of the heat received by the engine was converted
into work?
15. A gas engine receives 20,000 B.t.u. of heat in the form of fuel and
during the same period 3,11 2,000 footpounds of work are developed. What
is the thermal efficiency of the engine?
16. It is claimed that a certain motor generates 300,000 footpounds of
work per hour and during this period receives 400 B.t.u. of heat in the form
of fuel. Are such results possible?
17. An oil engine uses 0.74 lb. of fuel per b.h.p. per hour. Calculate the
thermal efficiency of this engine if the oil has a calorific value of 18,600 B.t.u.
per lb.
18. An engine receives 200 B.t.u. of heat per minute and exhausts during
the same period 100 B.t.u. If no losses of heat occur within the cylinder,
(a) How many B.t.u. of heat are being transformed into work?
(b) What number of footpounds does this heat produce?
(c) What horse power is being developed?
(d) What is the thermal efficiency of the engine?
19. In the manufacture of certain explosives, acids are mixed with an
oxidizable substance. During the process the mixture must be constantly
agitated by a stirring mechanism to maintain uniform conditions, and the
PROBLEMS 13
temperature of the mixture must be kept below a certain predetermined
value to prevent explosion. If during the process of manufacture 5000
footpounds of work are delivered per minute to the agitator, and 1000 B.t.u.
are generated during the same period by the chemical reaction, how much
heat must be absorbed per hour to maintain the temperature of the mixture
constant?
20. Prove that one horse power developed for one hour is equivalent to
the consumption of 2545 B.t.u. of heat in the same period.
CHAPTER II
PROPERTIES OF PERFECT GASES
In the study of thermodynamics, the working substance, or
heat medium, through which the heat engine converts heat into
work, is in the condition of a perfect gas or vapor. The laws
governing the action of the two classes of substances differ. For
this reason the study of thermodynamics is divided into the
thermodynamics of gases and of vapors.
When the term " gas " is used it refers to what is more properly
called a perfect gas. A perfect gas may be defined as a fluid
which remains in the gaseous state when subjected to changes in
pressure or in temperature. Gases which are near their point
of condensation are not perfect gases. Oxygen, hydrogen, nitro
gen, air and carbon dioxide are practical examples of perfect
gases.
Vapors are fluids which are readily transformed into liquids by
a very moderate reduction in temperature or increase in pressure.
Common examples of vapors are
steam and ammonia.
Relation between Pressure, Vol
ume and Temperature of a Per
fect Gas. In practically all heat
engines, work is done by changes
of volume of a fluid, and the
amount of work performed depends
only on the relation of pressure to
volume during such change and
not at all on the form of the vessel
containing this fluid.
Figure 2 shows a vessel containing a perfect gas and surrounded
by a jacket filled with cracked ice. Its temperature will, there
14
Connection to Air Pump
WJUIMEWa
Fig
— Constant Temperature
Apparatus for Demonstrating
Relation between Pressure and
Volume of a Gas.
BOYLE'S LAW 1 5
fore, be at 32 F. This vessel has a tightly fitting piston P of
which the lower flat side has an area of one square foot. In the
position shown the piston is two feet from the bottom of the
vessel, so that the volume between the piston and the bottom of
the vessel is two cubic feet. The pressure on the gas is that due
to the piston and the weights shown. Assume this total weight
is 100 pounds and that the air pump connected to the top of the
vessel maintains a vacuum above the piston. Then the pressure
on the gas below the piston is 100 pounds per square foot. If now
the weights are increased to make the pressure on the gas 200
pounds per square foot the piston will sink down until it is only
one foot from the bottom of the vessel, provided the ice keeps a
constant temperature. If the temperature is not maintained
constant, because of the tendency of gases to expand with in
crease in temperature, it will be necessary to apply a total weight
greater than 200 pounds to reduce the volume to one cubic foot.
Similarly, if the weight on the gas were reduced to 50 pounds and
the vessel were made high enough, the lower side of the piston
would then be four feet from the bottom of the vessel.
Examination of the above figures shows that if the temperature
is constant the product of pressure and volume is a constant, and
in this particular case it is always equal to 200 footpounds.
These facts are expressed by Boyle's Law which can be stated as
follows :
Boyle's Law. If a unit weight of gas is compressed or ex
panded at constant temperature, the pressure varies inversely as
the volume, or the product of pressure and volume remains a
constant. Thus, if P h V± are the initial pressure and volume and
P 2 , V 2 the final pressure and volume,
PiV 1 = P 2 V 2 . (14)
The laws of thermodynamics dealing with volume and pressure
changes corresponding to temperature variations may be stated
as follows:
(1) Under constant pressure the volume of a given mass of
gas varies directly as the absolute temperature.
16 PROPERTIES OF PERFECT GASES
(2) Under constant volume the absolute pressure of a given
mass of gas varies directly as the absolute temperature.
These fundamental principles, often called GayLussac's or
Charles* Laws, may also be stated thus:
V T
With pressure constant, —  = — , (15)
V 2 T 2
p j»
With volume constant, —  = — , (16)
"2 1 2
where V\ and V 2 are respectively the initial and final volumes,
Pi and P 2 are the initial and final absolute pressures, and 7\ and
T 2 are the absolute temperatures corresponding to the pressures
and volumes of the same subscripts.
The following problem shows applications of Charles' laws:
A gas has a volume of 2 cubic feet, a pressure of 14.7 pounds per
square inch absolute and a temperature of 6o° F.
(a) What will be the volume of this gas if the temperature is
increased to 120 F., the pressure remaining constant?
(b) What will be the pressure if the temperature is increased
as in (a) but the volume remains constant?
Solution, (a) Since the pressure remains constant and the
substance is a gas, the volume varies directly as the absolute
temperature.
Letting Vi and T\ be the initial conditions and V 2 and T 2 be
the final conditions, then
60 + 460
120 + 460'
V 2 = 2.23 cubic feet.
(b) Since the volume remains constant,
Pi T\ 14.7 60 + 460
or
V,
T,
2
or
v 2
T 2
v 2
P 2 T 2 P 2 120 + 460*
P 2 = 16.39 pounds per square inch absolute.
Combination of Boyle's and Charles' Laws. Equations (14),
(15), and (16) cannot often be used as they stand, because it
COMBINATION OF BOYLE'S AND CHARLES' LAWS 17
does not often happen that any one of the three variables (P, V
and P) remains constant. A more general law must be developed,
therefore, allowing for variations in all of the terms P, V and P.
This is accomplished by combining the above equations.
Assume a pound of gas of which the initial conditions of pres
sure, volume and temperature are represented by P h Vi and T 1}
while the corresponding final conditions are given by P 2 , V 2 and
P 2 . The first step is in changing the volume from Vi to V 2 and
the pressure from Pi to some intervening pressure Pi while the
temperature Pi remains constant. This change can be expressed
by Boyle's law (equation 14).
With constant temperature (Pi),
from which, by solving,
V, Pi
v 2 = pT (i7)
Pi = ^ ( l8 )
where Pi is the resulting pressure of the gas when its volume is
changed from Vi to V 2 , with the temperature remaining constant
at Pi.
The second step is in the change in pressure from Pi to P 2 and
in temperature from Pi to T 2 , while the volume remains con
stant at V 2 . This step is expressed as follows:
With constant volume (T 2 ),
Pi Pi / \
T, = T,' (l9)
which may be written
D P 2 T 2 , x
P2 = ~y— (20)
Substituting now the value of Pi from (18) in (20), we have
P > = F5T.' (2l)
1 8 PROPERTIES OF PERFECT GASES
which may be arranged to read,
P1V1 P2V2
(22)
The following problem shows the application of equation (22):
A quantity of air at atmospheric pressure has a volume of
2000 cubic feet when the barometer reads 28.80 inches of mercury
and the temperature is 40 C. What will be the volume of this
air at a temperature of o°C. when the barometer reads 29.96
inches of mercury?
Solution. Volume, pressure and temperature vary in this case
as in the following equation,
P1F1 P 2 V 2
then
T x T 2
Letting P h Vi, T\ = initial conditions,
P2, V 2) T 2 = final conditions,
28.80 X 2000 29.96 X V 2
40 + 273 o + 273
V 2 = 1676 cubic feet.
Now, since P 2 , V 2 and T 2 in equation (22) represent any
simultaneous condition of the gas, we may also write the follow
ing more general relations:
PiVx P 2 V 2 P 3 V
= a constant, (23)
2\ T 2 T 3
and, therefore,
PV = RT, (24)
PV
where R is the " gas constant," and is equal to — 
The Law of Perfect Gases. Equation (24) expresses the law
connecting the relation between pressure, volume and tempera
ture of a perfect gas. In this equation V is the specific volume,
or the volume occupied by a unit weight of a gas at the absolute
pressure P and absolute temperature T; R is the gas constant
THE LAW OF PERFECT GASES 19
in footpounds and depends on the density of the gas and on the
units of measurement adopted.
By means of equation (24) if the pressure, volume and tem
perature of a gas for one given condition are known, the value of
R can be determined.
Example 1. If the volume of air at freezing point and atmos
pheric pressure is 12.39 cubic feet per pound, calculate the value
of R in the English units.
Solution.. R = M* = I4 7 X 144 X 12.39 _ 5 .
To 32 + 460
Example 2. If one pound of air occupies 5 cubic feet at a
temperature of 200 F., find the corresponding pressure.
Solution. Using the value of R for air as calculated above
„ RTi ^.3 (460 + 200) ,
Px = — — = °° ° v ^  = 7040 pounds
Vi 5
per square foot absolute, which is the same as 34.2 pounds per
square inch gage pressure.
In calculating R by equation (24) care must be taken not to
confuse the units of measurement. It must also be remembered
that the method, as illustrated in the above examples, gives the
value of R for one pound of air, or for one unit weight of the gas
in question; for w pounds, the value of the constant is wR, or:
PV = wRT, (25)
where P = absolute pressure in pounds per square foot,
V = volume in cubic feet,
w = weight of gas in pounds,
R = the "gas constant" for one pound of gas in foot
pound units,
T = the absolute temperature in Fahrenheit degrees.
This equation is applicable to any perfect gas within the limits
of pressure and temperature employed in common engineering
practice. The " thermodynamic" state of a gas is known when
its pressure, volume, temperature, weight and composition are
known; when any four of these quantities are known the fifth can
be found by equation (25).
20 PROPERTIES OF PERFECT GASES
Heat and Its Effect Upon a Gas. In equation (12) it was
shown that in general the effect of adding heat upon a substance
was to increase the intrinsic energy and to overcome the external
resistances producing work. This law can be stated as follows:
Heat supplied = increase in intrinsic energy + the external
work done.
The converse of this statement is equally true. The heat
abstracted from a gas equals the decrease in intrinsic energy plus
the negative work done.
External Work. The external work or the work done by a
gas in its expansion is represented graphically by Fig. 3. The
area under the expansion line BC
^/w//w//m;w;mm//m/z
Pi
i
1
I
•
1
1
1
B
C
Sm
3
!/J
Cfl
V
u
Qui
is proportional to the work done
"^ in the expansion. If the initial
condition of the gas at B as re
gards pressure and volume is rep
resented by Pi and Vi and the
final condition at C by Pi and V 2
(expansion being at constant pres
Vi V 2 sure), the force P moved through
Volume / N
a distance of abscissas (k 2 — m)
Fig. 3. External Work of Expansion. is & measure of the WQrk done>
Obviously the area under the line BC divided by the horizontal
length (V 2 — Vi) is the average value of the force P. If, further,
and in general, we represent the area under BC by the symbol A,
then we can write,
— — = average value of P,
V 2 — ML
whether or not P is constant. And also,
Work done = A T _ X (7 2  Fi) = A.
The same principle applies whether the line BC is a straight line,
as shown in Fig. 3, or a very irregular curve, as will be shown
later.
INTERNAL ENERGY 21
Internal Energy. The heat energy possessed by a gas or
vapor, or the heat energy which is contained in a gas or vapor
in a form similar to " potential " energy, is called its internal
energy. It is also called intrinsic energy, since it may be said to
" reside " within the substance and has not been transferred
to any other substance. Thus, an amount of heat added to a
substance when no work is performed is all added to the internal
energy of that substance. On the other hand, when heat is added
while work is being performed, the internal energy is increased
only by the difference between the heat added and the work
done.
Internal energy may also be denned as the energy which a gas
or vapor possesses by virtue of its temperature, and for one
pound of gas may be expressed as follows :
Internal energy '= C v dT (in B.t.u.),
where T is the absolute temperature and C v the specific heat at
constant volume. From the paragraph on specific heat it will be
remembered that C v takes into account only that heat required
to raise the temperature, since under constant volume conditions
no external work is done; and, therefore, in dealing with internal
energy, C v is always used.
Increase in internal energy in B.t.u. (for one pound of a gas)
= C v (r 2  2\). (26)
Joule's Law. In the case of ideally perfect gases, such as
thermodynamic equations must deal with, it is assumed, when
a gas expands without doing external work and without taking
in or giving out heat (and, therefore, without changing its stock
of internal energy), that its temperature does not change. It was
for a long time supposed that when a gas expanded without doing
work, and without taking in or giving out heat, that its tempera
ture did not change. This fact was based on the famous experi
ments of Joule. Later investigations by Lord Kelvin and Linde
have shown that this statement is not exactly correct as all known
gases show a change in temperature under these conditions. This
change in temperature is known as the " JouleThomson" effect.
22 PROPERTIES OF PERFECT GASES
Relation of Specific Heats and the Gas Constant. If heat is
added at constant pressure, then
Q = wC P (T 2  7\). (27)
Also, by equation (26), the increase in internal energy when
heat is added
= wC,(T 2  7\). (28)
External work = P (V 2 — Vi) footpounds
 P (F2 7 Fl) B.t.u. (2 9 )
778 yy
Since the heat added = increase in internal energy + external
work,
wc p (t 2  ro = wc v (t 2  r x ) + ISXi^JA. ( 3o)
778
By equation (25),
P 2 V 2 = wRT 2 and P X V X = wRTl
Substituting these values in (30),
wC p (T 2 
 Ti) = wC v (T 2 — Ti) + w —
778
 (31)
Simplifying,
C = C 4 •
(32)
C P C =—= AR.
778
(33)
Equation (33) shows that the difference between the two
specific heats is equal to the gas constant R, which when measured
in footpounds represents the external work done by one pound
of a gas when its temperature is increased by one degree under
constant pressure.
Example. The specific heat of air at constant pressure (C p )
is 0.2375 B.t.u. and R = 53.3 footpounds. Calculate the spe
cific heat at constant volume (C») .
Solution.
C v = C p — AR = 0.2375 — ^ = 0.1690 B.t.u.
778
RATIO OF THE TWO SPECIFIC HEATS
2 3
Ratio of the Two Specific Heats (^r). The constant repre
senting the ratio of the two specific heats of a perfect gas is
represented by 7 where
n/ = — = J ( \
7 C v r AR AR K34)
Example. Calculate the value of 7 for air. C v = 0.2375 B.t.u.
R = 53.3 footpounds.
Solution. 7 = r— = = 1.40s.
j AR T 533
C p 778 x 0.2375
The constant designating the relation between the specific
heats of air is often designated by k instead of 7.
Values of the Specific Heats. Regnault after carrying on
experiments on the specific heat of hydrogen, oxygen, air and
•carbon dioxide concluded that all gases have a constant specific
heat under varying pressures and temperatures. Recent experi
ments ' tend to show that the specific heats of substances vary
with the pressure and temperature. The variability in the values
of the specific heats does not influence to any very great extent
most thermodynamic computations.
In the application of thermodynamics to internal combustion
engines the exact values of the specific heats are of considerable
importance.
Tables 1 and 2 in the appendix give constants for various gases.
PROBLEMS
1. Air at constant pressure with an initial volume of 2 cu. ft. and temper
ature of 6o° F. is heated until the volume is doubled. What is the resulting
temperature in degrees Fahrenheit?
2. Air is cooled at constant volume. The initial pressure is 30 lbs. per
24 PROPERTIES OF PERFECT GASES
sq. in. absolute and the initial temperature is ioi° F. The final condition
has a temperature of 50 F. What is the final pressure?
3. One pound of hydrogen is cooled at constant pressure from a volume
of 1 cu. ft. and temperature of 300 F. to a temperature of 6o° F. What is
the resulting volume?
4. A tank whose volume is 50 cu. ft. contains air at 105 lbs. per sq. in.
absolute pressure and temperature of 8o° F. How many pounds of air does
the tank contain?
5. An automobile tire has a mean diameter of 34 inches and a width of
4 inches. It is pumped to 80 lbs. per sq. in. gage pressure at a temperature
of 6o° F.; atmospheric pressure 14.6 lbs. per sq. in. absolute.
(a) How many pounds of air does the tire contain?
(b) Assuming no change of volume, what would be the gage pressure
of the tire if placed in the sun at ioo° F.?
6. A gas tank is to be made to hold 0.25 lb. of acetylene when the pres
sure is 250 lbs. per sq. in. gage, atmospheric pressure 14.4 lbs. per sq. in.
absolute, and the temperature of the gas 70 F. What will be its volume in
cubic feet?
7. A quantity of air at a temperature of 70 F. and a pressure of 15 lbs.
per sq. in. absolute has a volume of 5 cu. ft. What is the volume of the
same air when the pressure is changed at constant temperature to 60 lbs.
per sq. in. absolute?
8. How many pounds of air are required for the conditions in problem 7?
9. The volume of a quantity of air is 10 cu. ft. at a temperature of 6o° F.
when the pressure is 15 lbs. per sq. in. absolute. What is the pressure of
this air when the volume becomes 60 cu. ft. and the temperature 6o° F.?
10. How many pounds of air are required for the conditions in problem 9?
11. A tank contains 200 cu. ft. of air at a temperature of 6o° F. and under
a pressure of 200 lbs. per sq. in. absolute.
(a) What weight of air is present?
(b) How many cubic feet will this air occupy at 14.7 lbs. per sq. in.
absolute and at a temperature of ioo° F.?
12. The volume of a quantity of air at 70 F. and at a pressure of 14.2 lbs.
per sq. in. absolute is 20 cu. ft. What is the temperature of this air when
the volume becomes 5 cu. ft. and the pressure 80 lbs. per sq. in. absolute?
13. If the specific heat of carbon dioxide under constant pressure C v is
0.2012 and the value of R is 35.10, find the value of the specific heat under
constant volume C v .
14. How many B.t.u. are required to double the volume of one pound of
air at constant pressure from 50 F.
15. A tank filled with 200 cu. ft. of air at 15 lbs. per sq. in. absolute and
6o° F. is heated to 150 F.
PROBLEMS 25
(a) What will be the resulting air pressure in the tank?
(b) How many B.t.u. will be required to heat the air?
16. A tank contains 200 cu. ft. of air at 6o° F. and 40 lbs. per sq. in. abso
lute. If 500 B.t.u. of heat are added to it, what will be the resulting pres
sure and temperature?
CHAPTER III
EXPANSION AND COMPRESSION OF GASES
The equation of the perfect gas in the form PV = wRT for the
expansion or compression of gases has three related variables,
(i) pressure, (2) volume and (3) temperature. For a given
weight of gas with any two of these variables known the third is
fixed. As regards the analysis of the action of heat engines, the
pressure and volume relations are most important, and graphical
diagrams, called pressurevolume or PV diagrams, are frequently
needed to assist in the analysis. The indicator diagram is a pres
surevolume diagram drawn autographically by the mechanism
of an engine indicator. A pressure volume diagram is shown
in Fig. 4, in which the vertical scale of coordinates represents
pressures and the horizontal,
^ volumes. Assume that the pres
'^'m 1 ^ V3 7 Vl ~ v * sure and volume of a pound of a
« gas are given by the coordinates
P and Vi, which are plotted in
the middle of the diagram. It
& 24 6 8 io will be assumed further that
Volume, eu. ft.
the pressure remains constant in
Fig. 4. — Diagram of Expansion and , , , i • 1 • 1
~ . . r, , . t> the changes to be indicated.
Compression at Constant Pressure. to
Now if the gas is expanded until
its volume becomes V 2 , then its condition as regards pressure
and volume would be represented by P Vi. If, on the other hand,
the gas had been compressed while a constant pressure was
maintained, its final condition would be represented by the point
PVz to the left of PV\. Similarly, any line whether straight
or curved extending from the initial condition of the gas at PV\
will represent an expansion when drawn in the direction away
26
CD
EXPANSION AT CONSTANT PRESSURE 27
from the zero of volumes and will represent a compression when
tending toward the same zero.
It has been shown (page 20) that areas on such diagrams repre
sent the product of pressure and volume, and, therefore, work or
energy. Thus in Fig. 4 the area under the curve PV 3 to PV 2
represents on the scales given 100 (pounds per square foot) X
(9 — 1) cubic feet or 800 footpounds irrespective of whether it is
an expansion or a compression from the initial condition.
Most of the lines to be studied in heat engine diagrams are
either straight or else they can be exactly or approximately
represented by an equation in the form
PV n = a, constant, (35)
where the index n, as experimentally determined, has varying
numerical values, but is constant for any one curve. When
the lines of the diagram are straight the areas of simple rectangles
and triangles need only be calculated to find the work done.
The two most common forms of curves to be dealt with in ex
pansions are (1) when there is expansion with addition of heat
at such a rate as to maintain the temperature of the gas con
stant throughout the expansion. Such an expansion is called
isothermal. The other important kind of expansion (2) occurs
when work is done by the gas without the addition or abstraction
of heat. To do this work some of the internal heat energy
contained in the gas must be transformed in proportion to the
amount of work done. Such an expansion is called adiabatic.
The following problems show the application of the foregoing
principles to various types of expansions:
1. Expansion at Constant Pressure. One pound of air having
an initial temperature of 6o° F. is expanded to ioo° F. under
constant pressure. Find
(a) External work during expansion;
(b) Heat required to produce the expansion.
Solution. The heat added equals the increase in internal
energy plus the external work done. In solving for the heat
added or required during any expansion it is only necessary to
28 EXPANSION AND COMPRESSION OF GASES
find the external work (which is equal to the area under the ex
pansion curve) and add to it the heat needed to increase the
internal energy.
The external work = W = P l (V 2 V 1 ). (36)
Its equivalent is: wR (T 2 — 7\). (37)
W = 1 X 53.3 [(100 f 460) — (60 + 460)] = 2i32ft.lbs.
The increase in internal energy:
I = wC v (T 2  TO (38)
/ = I X 0.169 [(100 + 460) — (60 + 460)]
= 6.76 B.t.u.
Heat required (Q) = 6.75 H — = 9.50 B.t.u.
778
Another method of computing the heat required to produce
the expansion is:
e = ^c P (r 2 r 1 ). (39)
Q = 1 x 0.237 [( IO ° +.460) — (60 + 460)]
= 9.50 B.t.u. approximately.
2. Expansion at Constant Volume. One pound of air having
an initial temperature of 6o° F. is heated at constant volume until
the final temperature is 120 F. Find:
(a) External work;
(b) Heat required.
»
Solution.
Heat added (Q) = increase in internal energy f external work.
External work = o.
Then
Heat added = increase in internal energy + o
= wC v (T 2  Ti) + o
= 1 X 0.169 [(100 + 460) — (60 + 460)] + o
= 6.76 B.t.u.
3. Expansion and Compression at Constant Temperature
(Isothermal). In an isothermal expansion or compression the
ISOTHERMAL EXPANSION AND COMPRESSION
29
temperature of the working substance is kept constant through
out the process. The form of the isothermal curve on pressure
volume coordinates depends upon the substance. In the case
of perfect gases Boyle's Law (equation 14) applies and we have:
PV = C = a constant.
(40)
Equation (40) is that of a rectangular hyperbola. It is the
special case of the general equation PV n = constant (35), in
which the index n = 1 and is represented by Fig. 5.
Volume dV
Fig. 5. — Work done during Isothermal Expansion and Compression.
The external work performed is shown graphically by the
shaded area under the curve between A and B (Fig. 5). The
two vertical lines close together in the figure are the limits of a
narrow closely shaded area and indicate an infinitesimal volume
change dV. Work done during this small change of volume is,
the
JW = P dV,
and for a finite change of volume of any size as from V\ to V 2
the work done, W (footpounds), is:
W
PdV.
(41)
30 EXPANSION AND COMPRESSION OF GASES
For the integration of this form it is necessary to substitute P
in terms of V. Assume that P and V are values of pressure
and volume for any point on the curve of expansion of a gas of
which the equation is
PV = C.
Then P = £•
Substituting this value of P in equation (41),
tv *dV
Vl v
W = C (log. V 2  loge FO. (42)
Since the initial conditions of the gas are Pi and Vi, we have
PV = C = PxFi,
and substituting this value of C in equation (42), we obtain
W = PxFi (log, F 2  log e 70
or T7 = P1V1 loge :==? (in footpounds). (43)
r 1
Units of weight do not enter in equations (42) and (43). For
a certain weight of gas under the same conditions, since
Vo P
P1V1 = wRT (in footpounds) and — = =^>
Vi P%
then the work for w pounds is :
V P
W = wRT loge — = wRT log e 5^ (in footpounds). (44)
V I P2
. V*
Often the ratio — is called the ratio of expansion and is rep
V\
resented by r. Making this substitution we have,
W = wRT log s r. (45)
Equations (42), (43), (44) and (45) refer to an expansion from
P1V1 to P2V2. If, on the other hand, the work done is the result
of a compression from P1V1 to P3V3 the curve of compression
ISOTHERMAL EXPANSION AND COMPRESSION 31
would be from A to C and the area under it would be its graphical
representation. Equations (43), (44) and (45) would represent
the work done for compression the same as for expansion, except
that the expression would have a negative value; that is, work is
to be done upon the gas to decrease its volume.
The isothermal expansion or compression of a perfect gas
causes no change in its stock of internal energy since the temper
ature T is constant. During such an expansion the gas must
take in an amount of heat just equal to the work it does, and
conversely during an isothermal compression it must reject an
amount of heat just equal to the work spent upon it. This
quantity of heatQ (in B.t.u.) is, from equation (45),
n wRT , V 2 t ,,
Q = w ge v; (46)
The following problem shows the application of the foregoing
formulas to isothermal expansions:
Air having a pressure of 100 pounds per square inch absolute
and a volume of 1 cubic foot expands isothermally to a volume
of 4 cubic feet. Find
(a) External work of the expansion;
(b) Heat required to produce the expansion;
(c) Pressure at end of expansion.
Solution, (a) Since the expansion is isothermal,
External work, W = Pi7i log e * ^
= IOO X 144 X I x 1.3848
= 19,940 footpounds.
(b) Since the heat added equals increase in internal energy
plus external work, and since the temperature remains constant
(requiring therefore no heat to increase the internal energy), the
internal energy equals zero and the heat added equals the work
done.
* 2.3 X log base 10 = log base e. Tables of natural logarithms are given in the
appendix.
32
EXPANSION AND COMPRESSION OF GASES
Then: Heat added = external work = 9>94Q = 25.6 B.t.u.
778
(c)
then
Since
P1V1
100 X 1
P 2
P 2 F 2 ,
P 2 x 4,
25 pounds per square inch absolute.
If a gas expands and does external work without receiving a
supply of heat from an external source, it must derive the amount
of heat needed to do the work from its own stock of internal
energy. This process is then necessarily accompanied by a low
ering of temperature and the expansion obviously is not iso
thermal.
4. Adiabatic Expansion and Compression. In the adiabatic
mode of expansion or compression the working substance neither
receives nor rejects heat as it expands or is compressed. A
curve which shows the relation of pressures to volumes in such
a process is called an adiabatic line (see Fig. 6). In any adiabatic
Volume
Fig. 6. — Isothermal and Adiabatic Expansion Lines.
process the substance is neither gaining nor losing heat by
conduction, radiation or internal chemical action. Hence the
work which a gas does in such an expansion is all done at the
expense of its stock of internal energy, and the work which is
done upon a gas in such a compression all goes to increase its
internal energy. Ideally adiabatic action could be secured by
a gas expanding, or being compressed, in a cylinder which in all
parts was a perfect nonconductor of heat. The compression of
a gas in a cylinder is approximately adiabatic when the process is
very rapidly performed, but when done so slowly that the heat
ADIABATIC EXPANSION AND COMPRESSION 33
has time to be dissipated by conduction the compression is more
nearly isothermal. Fig. 6 shows on a pressurevolume diagram
the relation between an isothermal and an adiabatic form of ex
pansion or compression from an initial condition P1V1 at A to
final conditions at B and C for expansions, and at D and E for
compressions.
In order to derive the pressurevolume relation for a gas
expanding adiabatically, consider the fundamental equation
(page 20):
Heat added = increase in internal energy + external work, or:
.<2 = wK v (T 2  TO + PdV (footpounds), (47)
where K v is the specific heat in footpound units, i.e., 778 C v .
In the adiabatic expansion no heat is added to or taken away
from the gas by conduction or radiation, and, therefore, the left
hand member of the above equation becomes zero. Furthermore,
since equation (25) can always be applied to perfect gases, the
following simultaneous equations may be written:
o = wK v dT + P dV, (48)
PV = wRT. (49)
When P, V and T vary, as they do in adiabatic expansion,
equation (49) may be written as follows:
PdV+ VdP = wRdT, (50)
and
PdV + VdP
dT =
wR
Substituting the value of dT in (48),
n^T/ , v PdV + VdP , ,
PdV + wK v l — = o. (51)
wR
RPdV + K V P dV + K V V dP = o.
34 EXPANSION AND COMPRESSION OF GASES
To separate the variables divide by PV:
W 4. K dV + K dF
p dV dV dP . .
R — + K, — + K, — = o. (52)
Collecting terms,
(i? + K.) ^ + tf. ^ = o.
Integrating,
(R + K v ) log e F + A% log e P = a constant = c. (53)
*±*'log a F + log.P = C .
Ay
ig + gp
lo ge PF * =c, (54)
since from equation (^),
R + K v = Kp y
where K v and A% are respectively the specific heats at constant
pressure and at constant volume in footpound units, then equa
tion (54) becomes:
From equation (34),
\og & PV Kv =c.
therefore :
r~ = y > or k~ = 7 >
PV 7 = a constant. (55)
Following the method used for obtaining an expression for
the work done in isothermal expansion (equation 43), the work
done, W (in footpounds), for a change of volume from Vi to F 2 ,
Jf*Vi
PdV. (56)
W
V 2
For purposes of integration, P can be substituted in terms of
V as outlined below. In the general expression PV n = c,sl con
stant (see equation 35), where P and V are values of pressure and
volume for any point on the curve of expansion of a gas of which
ADIABATIC EXPANSION AND COMPRESSION 35
the initial condition is given by the symbols Pi and V h we can
then write,
p = — (57)
And substituting (57) in (56),
r v~ n+1 Y 2
L n + ij 7l
= 1 i„ J (59)
Since PV n = c = PiV? = P 2 V 2 n , we can substitute for c in
(59) the values corresponding to the subscripts of V as follows:
P 2 V 2 n V 2 l ~ n  PiF/% 1 "
W =
w =
1 — n
P 2 V 2  P 1 V 1
)
1 — n
or W = ^ ^^ (footpounds). (60)
n — 1
Since PF = wPP,
PT = wR (Tl ~ T2)  (footpounds). (61)
n — 1
Equations (60) and (61) apply to any gas undergoing expan
sion or compression according to PV n = a constant. In the case
of an adiabatic expansion of a perfect gas n = 7 (see equation
55).
Change of Internal Energy During Adiabatic Processes.
Since in an adiabatic expansion no heat is conducted to or away
from the gas, the work is done at the expense of the internal
energy and, therefore, the latter decreases by an amount equiva
lent to the amount of work performed. This loss in internal
energy is readily computed by equation (60) or (61). The
result must be divided by 778 in order to be in B.t.u.
During an adiabatic compression the reverse occurs, i.e., there
36 EXPANSION AND COMPRESSION OF GASES
is a gain in internal energy and the same formulas apply, the
result coming out negative, because work has been done on the
gas.
Relation between Volume, Pressure and Temperature in
Adiabatic Expansion of a Perfect Gas. Since P, V and T vary
during an adiabatic expansion, it will be necessary to develop for
mulas for obtaining these various quantities. It will be remem
bered that equation (25), applies to perfect gases at all times.
Therefore, in the case of an adiabatic expansion or compression
we can write the two simultaneous equations:
■££.££•• (A)
±1 1 2
p x v x y = p 2 v 2 y . (B)
By means of these two equations we can find the final condi
tions of pressure, volume and temperature, having given two
initial conditions and one final condition.
For instance, having given V\, V 2 and T h to find T 2) divide (A )
by {B), member for member. Then
V 1 V 2
TxV^ T 2 V 2 y
T 2 = Vjv
T 2 = T,'
or
In like manner the following formulas can be obtained:
(63)
p
rM y . («4)
ADIABATIC EXPANSION AND COMPRESSION 37
P* = Pi (ffi*. (65)
V, = V 1 (^p, (66)
1
Fs=Fl (S) 71 ' (67)
It should be noted that the above formulas can be used for
any expansion of a perfect gas following PF" = a constant,
provided y in the formulas is replaced by n.
It is also to be noted that these equations can be used for
any system of units so long as the same system of units is
employed throughout an equation.
There are many cases of expansions which are neither adiabatic
nor isothermal and which are not straight lines on PV diagrams.
It will be observed from the equations in the discussion of the
internal work done by an expanding gas and for the change of
internal energy, that if in the general equation PV n = a. constant
the exponent or index n is less than y, the work done is greater
than the loss in internal energy. In other words for such a
case, the expansion lies between an adiabatic and isothermal and
the gas must be taking in heat as it expands. On the other
hand, if n is greater than y the work done is less than the loss
of internal energy.
Example. Given a quantity of pure air in a cylinder at a tem
perature of 6o° F. (7\ = 460 + 60 = 520 degrees absolute) which
is suddenly (adiabatically) compressed to half its original volume.
Then — 1 = — , and taking y as 1.405, the temperature immediately
V 2 1
after compression is completed, T 2 , is calculated by equation (62)
as follows:
/FA 71 /2V 405  1
T 2 = Ti I y) = 520 I J = 520 X 2 0405 = 688° absolute,
or t 2 in ordinary Fahrenheit is 688 — 460 or 228 degrees.
38 EXPANSION AND COMPRESSION OF GASES
The work done in adiabatic compression of cAe pound of this
air is calculated by equation (61):
w= wRjT.T,) = 53.3(520688) = 533 (168) = _ 22jIIQ
71 1.405 1 0.405
footpounds per pound of air compressed. The negative sign
means that work has been done on the gas or the gas was com
pressed. If the sign had been positive it would have indicated
an expansion.
As the result of this compression the internal energy of the gas
has been increased by — B.t.u., but if the cylinder is a con
_ 778
ductor of heat, as in practice it always is, the whole of this heat
will become dissipated in time by conduction to surrounding
air and other bodies, and the internal energy will gradually
return to its original value as the temperature of the gas comes
back to the initial temperature of 6o° F.
PROBLEMS
1. Calculate the heat required to produce a temperature change of 50 F.
in three pounds of air at constant pressure.
2. How many footpounds of work are done by 2 lbs. of air in expand
ing to double its volume at a constant temperature of ioo° F.?
3. Three pounds of air are to be compressed from a volume of 2 to 1 cu. ft.
at a constant temperature of 6o° F. How many B.t.u. of heat must be
rejected from the air?
4. An air compressor has a cylinder volume of 2 cu. ft. If it takes air
at 14.4 lbs. per sq. in. absolute and 70 F. and compresses it isothermally to
100 lbs. per sq. in. absolute, find
(a) Pounds of air in cylinder at beginning of compression stroke.
(b) The final volume of the compressed air.
(c) The footpounds of work done upon the gas during compres
sion.
(d) The B.t.u. absorbed by the air in increasing the internal energy,
(c) The B.t.u. to be abstracted from the cylinder.
5. A cubic foot of air at a pressure of 150 lbs. per sq. in. gage expands
isothermally until its pressure is 50 lbs per sq. in. gage. Calculate the
work done during this expansion.
PROBLEMS 39
6. Air at ioo lbs per sq. in. absolute and a volume of 2 cu. ft. expands
along an» = 1 curve to 25 lbs. per sq. in. absolute pressure. Find
(a) Work done by the expansion.
(b) Heat to be supplied.
7. A quantity of air at 100 lbs. per sq. in. absolute pressure has a tem
perature of 8o° F. It expands isothermally to a pressure of 25 lbs. per
sq. in. absolute when it has a volume of 4 cu. ft. Find (1) the mass of air
present, (2) Work of the expansion in footpounds, (3) Heat required in
B.t.u.
8. Air at 100 lbs. per sq. in. absolute pressure and 2 cu. ft. expands to
25 lbs. per sq. in. absolute adiabatically. What is the final volume?
9. One cubic foot of air at 6o° F. and a pressure of 15 lbs. per sq. in.
absolute is compressed without loss or addition of heat to 100 lbs. per sq.
in. absolute pressure. Find the final temperature and volume.
10. Two pounds of air are expanded from a temperature of 300 F. to
200 F. adiabatically. How many footpounds of work are developed?
11. A quantity of air having a volume of 1 cu. ft. at 6o° F. under a pres
sure of 100 lbs. per sq. in. absolute is expanded to 5 cu. ft. adiabatically^
Find the pounds of air present, the final temperature of the air and the
work done during this expansion.
12. Data the same as in Problem 4 but the compression is to be
adiabatic. Find
(a) The final volume of the compressed air.
(b) The final temperature of the compressed air.
(c) The footpounds of work required to compress this air.
(d) The B.t.u. absorbed by the air in increasing the internal
energy.
(e) The B.t.u. to be abstracted from the gas.
13. A pound of air at 32 F. and atmospheric pressure is compressed to
4 atmospheres (absolute). What will be the final volume and the work
of compression if the compression is (a) isothermal, (b) adiabatic?
14. Plot the curve PV n = C, when n = 1.35. Initial pressure is 460
lbs. per sq. in. gage, initial volume 0.5 cu. ft., and final volume 8 cu. ft.
15. Prove that the work of an adiabatic expansion can be expressed by
the formula:
"—[(in
CHAPTER IV
CYCLES OF HEAT ENGINES USING GAS
The Heat Engine Cycle. In the heat engine, the working sub
stance or heat medium undergoes changes in its physical proper
ties converting heat into mechanical work. The series of such
changes by the repetition of which the conversion of heat into
work takes place forms the heat engine cycle.* The heat engine
cycle usually consists of four events which are: heating, cooling,
expansion, and compression.
i. THE CARNOT CYCLE
Very important conclusions regarding theoretically perfect
heat engines are to be drawn from the consideration of the action
of an ideal engine in which the working substance is a perfect
gas which is made to go through a cycle of changes involving
both isothermal and adiabatic expansions and compressions.
This ideal cycle of operations was invented and first explained in
1824 by Carnot, a French engineer. This cycle gave the first
theoretical basis for comparing heat engines with an ideally
perfect engine. The ideal Carnot cycle requires an engine,
illustrated in Fig. 7, which consists of the following parts:
(1) A piston and cylinder, as shown in Fig. 7, composed of
perfectly nonconducting material, except the cylinderhead (left
hand end of the cylinder) which is a good conductor of heat. The
space in the cylinder between the piston and the cylinder head is
occupied by the working substance, which can be assumed to be
a perfect gas.
* A thermodynamic machine performing a cycle in which heat is changed into
work is called a heat engine, and one performing a cycle in which heat is trans
ferred from a medium at a low temperature to one at a higher temperature is
called a refrigerating machine.
40
THE CARNOT CYCLE
41
(2) A hot body H of unlimited heat capacity, always kept at a
temperature TV.
(3) A perfectly nonconducting cover N.
(4) A refrigerating or cold body R of unlimited heat receiving
capacity, which is kept at a constant temperature T% (lower
than Ti).
H
N
Fig. 7. — Apparatus and Diagram Illustrating a Reversible (Carnot) Cycle.
It is arranged that H, N or R can be applied, as required, to
the cylinder head. Assume that there is a charge of one pound
of gas in the cylinder between the piston and the cylinder head,
which at the beginning of the cycle, with the piston in the posi
tion shown, is at the temperature T h has a volume V a , and has a
pressure P a . The subscripts attached to the letters V and P
refer to points on the pressure volume diagram shown in Fig. 7.
42 CYCLES OF HEAT ENGINES USING GAS
This diagram shows, by curves connecting the points a, b, c and
d, the four steps in the cycle.
The operation of this cycle will be described in four parts as
follows:
(i) Apply the hot body or heater H to the cylinder head at
the lefthand side of the figure. The addition of heat to the gas
will cause it to expand isothermally along the curve ab, because
the temperature will be maintained constant during the process
at T\. The pressure drops slightly to P b when the volume be
comes Vb. During this expansion external work has been done
in advancing the piston and the heat equivalent of this work has
been obtained from the hot body H.
(2) Take away the hot body H and at the same time attach
to the cylinder head the nonconducting cover N. During this
time the piston has continued to advance toward the right,
doing work without receiving any heat from an external source,
so that the expansion of the gas in this step has been done at
the expense of the stock of internal energy in the gas along the
adiabatic curve be. The temperature has continued to drop in
proportion to the loss of heat to the value T 2 . Pressure is then
P c and the volume is V c .
(3) Take away the nonconductor N and apply the refriger
ator R. Then force the piston back into the cylinder. The
gas will be compressed isothermally at the temperature T 2 .
In this compression, work is being done on the gas, and heat is
developed, but all of it goes into the refrigerator R, in which the
temperature is always maintained constant at T 2 . This com
pression is continued up to a point d in the diagram, so selected
that a further compression (adiabatic) in the next (fourth)
stage will cause the volume, pressure and temperature to reach
their initial values as at the beginning of the cycle*
(4) Take away the refrigerator R and apply the nonconduct
ing cover N. Then continue the compression of the gas without
* Briefly the third stage of the cycle must be stopped when a point d is reached,
so located that an adiabatic curve (PV 7 = constant) drawn from it will pass
through the "initial" point a.
THE CARNOT CYCLE 43
the addition of any heat. It will be the adiabatic curve da.
The pressure and the temperature will rise and, if the point d has
been properly selected, when the pressure has been brought back
to its initial value P a the temperature will also have risen to its
initial value 2\. The cycle is thus finished and the gas is ready
for a repetition of the same series of processes comprising the
cycle.
To define the Carnot cycle completely we must determine
how to locate algebraically the proper place to stop the third
step (the location of d). During the second step (adiabatic ex
pansion from b to c) by applying equation (62), the following
temperature and volume relations exist:
Ti [Vol 7  1
T 2 lV b ] '
also for the adiabatic compression the fourth step can be sirni
lary stated,
Hence,
wr  wr
Simplifying and transposing, we have
V b V c
v. v, < 68)
V
— is the ratio of expansion r for the isothermal expansion in
V a
the first step of the cycle. This has been shown (Equation 68) to
be equal to —  in the isothermal compression in the third step in
order that the adiabatic compression occurring in the fourth
step shall complete the cycle.
A summary of the heat changes to and from the working
gas (per pound) in the four steps of the Carnot cycle is as fol
lows:
44 CYCLES OF HEAT ENGINES USING GAS
(ab). Heat taken in from hot body (by equation (45), in
footpounds) is:
RTl\0g e ^ (69)
• V a
(be). No heat taken in or rejected.
(cd). Heat rejected to refrigerator (by equation (45), in
footpounds) is:
V V *
RT 2 log, — =  RT 2 log e /• (70)
V c V d
(da). No heat taken in or rejected.
Hence, the net amount of work done, W, by the gas in this
cycle, being the mechanical equivalent (footpounds) of the
excess of heat taken in over that rejected, is the algebraic sum
of (69) and (70):
W = r(t 1 log, y a T, log, £) =R(T l  r 2 ) loge £• (71)
The thermal or heat efficiency of a cycle is defined as the ratio of
Heat equivalent of work done
Heat taken in
The heat equivalent of work done is, by equation (71),
ic^T^log,^,
and the heat taken in is, by equation (69),
The ratio above representing the efficiency E is:
r (z\  r 2 ) log. ^
E ^ = T ^ (72)
RT, log, £
V a
Log togpj.
REVERSIBLE CYCLES 45
The efficiency of the Carnot cycle as expressed by equation (72)
is the maximum possible theoretical efficiency which may be
obtained with any heat engine working between the temperature
limits 7\ and T 2 . This equation can be used as a standard for the
comparison of the efficiencies of actual heat engines working
between two temperature limits.
Reversible Cycles. A heat engine which is capable of dis
charging to the "source of heat" when running in the reverse
direction from that of its normal cycle the same quantity of
heat that it would take from this source when it is running
direct and doing work is said to operate with its cycle reversed,
or, in other words, the engine is reversible. A reversible heat
engine then is one which, if made to follow its indicator diagram
in the reverse direction, will require the same horse power to drive
it as a refrigerating machine as the engine will deliver when
running direct, assuming that the quantity of heat used is the
same in the two cases. An engine following Carnot's cycle is,
for example, a reversible engine. The thermodynamic idea of
reversibility in engines is of very great value because no heat
engine can be more efficient than a reversible engine when both
work between the same limits of temperature; that is, when
both engines take in the same amount of heat at the same higher
temperature and reject the same amount at the same lower
temperature.
It was first proved conclusively by Carnot that no other heat
engine can be more efficient than a reversible engine when both
work between the same temperature limits. To illustrate this
principle, assume that there are two engines A and B. Of these
let us say A is reversible and B is not. In their operation both
take heat from a hot body or heater H and reject heat to a refrig
erator or cold body R. Let Q H be the quantity of heat which
the reversible engine A takes in from the hot body H for each
unit of work performed, and let Q B be the quantity of heat per
unit of work which it discharges to the refrigerator R.
For the purpose of' this discussion, assume that the non
reversible engine B is more efficient than the reversible engine A.
46 CYCLES OF HEAT ENGINES USING GAS
Under these circumstances it is obvious that the engine B will take
in less heat than A and it will reject correspondingly less heat
to R per unit of work performed. The heat taken in by the
nonreversible engine B from the hot body H we shall designate
then by a quantity less than Q H , or Q H — X, and the heat rejected
by B to the refrigerator R by Q R — X. Now if the nonreversible
engine B is working direct (when converting heat into work)
and is made to drive the reversible engine A according to its
reverse cycle (when converting work into heat), then for every
unit of work done by the engine B in driving the reversible
engine A, the quantity of heat mentioned above, that is, Q H — X,
would be taken from the hot body H by the nonreversible
engine B and, similarly, the quantity of heat represented by Q H
would be returned to the hot body H by the reverse action of
the cycle of operations performed by A. This follows because
the engine A is reversible and it returns, therefore, to H, when
operating on the reverse cycle, the same amount of heat as it
would take in from H when working on its direct cycle. By
this arrangement the hot body H would be continually receiving
heat, in the amount represented by X for each unit of work
performed. At the same time the nonreversible engine B dis
charges to the refrigerator R a quantity of heat represented by
Q R — X, while the reversible engine A removes from the refriger
ator R a quantity represented by Q R . As a result of this last
operation the cold body will be losing continually per unit of work
performed a quantity of heat equal to X. The combined per
formances of the two engines, one working direct as a normal heat
engine and the other, according to its reverse cycle, as a com
pressor or what might be called a "heat pump/' gives a constant
removal of heat from the refrigerator R to the hot body H,
and as a result a degree of infinite coldness must be finally pro
duced in the refrigerator.
If we assume that there is no mechanical friction, this com
bined machine, consisting of a normal heat engine and com
pressor, will require no power from outside the system. For
this reason the assumption that the nonreversible engine B
REVERSIBLE CYCLES
47
can be more efficient than the reversible engine A has brought
us to a result which is impossible from the standpoint of expe
rience as embodied in the statement of the " Second Law of
Thermodynamics "; that is, it is impossible to have a selfacting
engine capable of transferring heat, infinite in quantity, from a
cold body to a hot body. We should, therefore, conclude that
no nonreversible engine, as B for example, can be more efficient
than a reversible engine A when both engines operate between
the same temperature limits. More briefly, when the source of
heat and the cold receiver are the same for both a reversible heat
engine and any other engine, then the reversible engine must
have a higher possible efficiency; and if both engines are reversible
it follows that neither can be more efficient than the other.
A reversible engine is perfect from the viewpoint of efficiency;
that is, its efficiency is the best obtainable. No other engine
than a reversible engine which takes in and discharges heat at
identical temperatures will transform into work a greater part of
the heat which it takes in. Finally, it should be stated as regards
this efficiency that the nature of the substance being expanded or
compressed has absolutely no relation to the thermal efficiency
as outlined above.
If an engine operating on Carnot's cycle is reversed in its action,
so that the same indicator diagram shown in Fig. 7 would be
traced in the opposite direction, the reversed cycle, when begin
ning as before at point a with a perfect gas at the temperature
Ti, will consist of the following stages:
(1) When the nonconductor N is applied and the piston is
advanced toward the right by the source of power performing
the reversed cycle, the gas will expand, tracing the adiabatic
curve ad, with constant lowering of temperature which at the
point d will be T 2 .
(2) When the nonconductor N is now removed, the refrig
erator R is applied, and the piston continues on its outward stroke.
The gas will expand isothermally at the constant temperature T 2 ,
tracing the curve dc. During this stage the gas is taking heat
from the refrigerator R.
48 CYCLES OF HEAT ENGINES USING GAS
(3) When the refrigerator R is removed and the nonconductor
N is again applied, which will be on the back stroke of the engine,
the gas will be compressed, and on the indicator diagram another
adiabatic curve cb will be traced. At the point b the temper
ature will be obviously T\.
(4) When the nonconductor N is removed and the hot body
H is again applied, with the compression continuing along the
isothermal curve ba, heat will be discharged to the hot body H,
while the temperature is maintained constant at 7\.
The cycle has now been traced in a reverse direction from the
beginning back to the starting point at a, and is now complete.
During this process an amount of work represented by the area
of the indicator diagram, equivalent in footpounds to
R log e —  (7\ — T 2 ) (see equation 71),
V a
has been converted into heat. First, heat was taken from the
refrigerator R, represented in amount by
RT 2 logo ■=£■>
V d
and second, heat was rejected to the hot body H in the amount
RTylog^ or 2?7\log3.
As in direct operation of Carnot's cycle no heat is given or lost
in the first and third stages outlined above. The algebraic sum
of these two quantities, remembering that — = =^ } gives the
V a V d
net amount of work done, W, on the gas, and, therefore, the
net amount of heat (footpound units) transferred from the cold
body R to the hot body H or,
W = RT 2 log e ^  RT, log, £ = R log c £ (7\  T 2 ). (73)
V a y a y a
HOTAIR ENGINE CYCLES 49
Since the result is the same as given by equation (72), although
opposite in sign on account of being work of compression, it
will be observed that in the reverse cycle the same amount
of heat is given to the hot body H as was taken from it in the
direct operation of the same cycle, and that the same amount
of heat is now taken from the refrigerator R as was in the other
case given to it.
2. HOTAIR ENGINE CYCLES
The hotair engine is a nonexplosive type of external combus
tion heat engine, the working substance being atmospheric air
which undergoes no change in its physical state. This type of
primemover was invented about 100 years ago. It is little used
on account of the difficulty in transmitting heat through the
metallic walls to the dry gas air. Then the hotair engine is
very bulky in proportion to its power. There are also difficul
ties in carrying out practically the theoretical cycles of operation,
on account of the rapid deterioration of the heatconducting sur
faces. The mechanical efficiency of the hotair engine is also low.
The advantages of the hotair engine are ease of operation and
safety. Also its speed being low, on account of the rate of heat
transmission, it is well suited for the driving of small pumps and
for other domestic uses where small powers are required and also
where the fueleconomy is a matter of minor importance.
Hotair engine cycles are divided into two groups : —
Group I. Externalcombustion hotair engines with a closed
cycle and constant volume temperature changes.
Group II. Externalcombustion hotair engines with an open
cycle and constantpressure temperature changes.
In the hotair engines an attempt was made to put in practice
a cycle of the ideal Carnot type with the addition of a regenera
tor. The regenerator was a storagebattery of heat, its function
being to absorb, store and return heat rapidly, replacing the
adiabatic expansion curves of the Carnot cycle by lines of con
stant volume in group I and by lines of constant pressure in
5o
CYCLES OF HEAT ENGINES USING GAS
group II. The regenerator consists of a chamber containing
strips of metal, coils of wire, or any other heatabsorbing material
arranged in such a manner as to present a very large surface to
the air passing through it.
The Stirling Engine Cycle. The Stirling engine belongs to
group I, the regenerator being so arranged that the pressure
chops with the temperature at such a rate as to keep the vol
ume constant. It is an externalcombustion engine and its
I
)
\>
V*
N&
^&>
^
a
v>
X r 4„
d
Volume
Fig. 8. — Stirling Engine Cycle.
cycle of operation is closed, the same air is used over and
over again, any loss by leakage being supplied by a small
forcepump. The engine consists, essentially, of a pair of dis
placer cylinders, a doubleacting working cylinder, a regenera
tor and a refrigerator. In the earlier forms the displacer cyl
inder had a double wall, the regenerator and refrigerator being
placed in the annular space surrounding the displacer cylinder.
In the latter forms the regenerator and refrigerator are arranged
separately from the displacer cylinder but in direct communi
cation with its top and bottom. A plunger works in the dis
placer cylinder, this plunger being filled with a nonconducting
material such as brickdust. The engine derives its heat by
STIRLING ENGINE CYCLE 5 1
conduction from a furnace which is placed beneath the displac
ing cylinder.
In the cycle represented by Fig. 8, the action of the engine
is carried out as follows:
1. The substance, air, having a volume V a , a pressure P a and
a temperature T a receives heat at constant volume by passing
through the regenerator, and receives also heat from the furnace.
As a result of the heat addition, its pressure is increased to P b , its
temperature to T b , this process being represented by the constant
volume line ab, Fig. 8. During this event the plunger is moving up.
2. The pressure under the working piston increases as a result
of the addition of heat and the expansion of the air follows,
producing the working stroke. This expansion is represented by
be and is an isothermal, the air receiving heat from the furnace
plates and the temperature remaining 7\.
3. The plunger descends, displacing the hot air and forcing the
same through the regenerator and refrigerator. As a result of
this, the temperature drops from T c to T& and the pressure is
decreased from P c to P d , the volume remaining constant as shown
by the line cd.
4. Due to this cooling effect the working piston descends com
pressing the air. This compression, represented by da, is iso
thermal, the air being cooled during the process by the regenera
tor and refrigerator to remain at the temperature TV
Referring to Fig. 8 of the Stirling cycle, the useful work W
is in footpounds,
W = (jP b V b log, y>)  (p d V d log, y)' (H)
The mean effective pressure is in pounds per square foot,
M.E.P. = W ■ (75)
V d — V a
The horse power developed can be represented by
H.P. = 5^ ( 7 6)
33,000
where N = revolutions (cycles) per minute.
52
CYCLES OF HEAT ENGINES USING GAS
Since the expansion and compression processes are isothermal,
the efficiency of the Stirling cycle is:
P b V b log e ^P d V d log ]
E
Vt
V a
P b V b log e
Vc
V b
(77)
Since V a = V b and V c = V d , also RTi = P b V b and RT 2 =
PdV d , equation (77) becomes,
E =
RTr loge — c  RT^loge =/
Vb V b
Vc
RT ± loge ^r
Vb
Tt T 2
(78)
The Ericsson Engine Cycle. This engine belongs to group II,
the regenerator changing the temperature at constant pressure.
This engine consists of five parts : a compressing pump, a receiver,
a regenerator, a refrigerator and a working cylinder. The ideal
a'
6
m
CL,
a
b
\Ti
•
NO* r ^
\Isothermal
^^sothermal •
r
d
Volume
Fig. 9. — The Ericsson Engine Cycle.
cycle for this engine is represented by Fig. 9, the order of events
being as follows :
Atmospheric air is drawn into the compressing pump as shown
by d'd. This air is compressed isothermally to a, during the
return stroke of the pump; it is then forced into a receiver as
shown by aa! . Thus d'daa' represents the pump cycle.
As the working cylinder begins its forward or up stroke, the
THE LENOIR ENGINE 53
compressed air is admitted into the working cylinder at constant
pressure. As this admission from a to b is through the regenera
tor, the entering air takes up heat increasing in temperature from
Ti to T\. As soon as the supply of compressed air is cut off,
the air expands isothermally along be to atmospheric pressure,
while heat is being supplied by a furnace at the bottom of the
working cylinder. During the return stroke of the working pis
ton, the air is discharged at constant pressure through the regen
erator, giving up its heat and is cooled to the temperature T 2 .
In Fig. 9, a' bed' is the cycle of the working cylinder, the net work
being represented by abed.
The Ericsson cycle has the same efficiency as the Stirling cycle,
the regenerator process being carried out at constant pressure
instead of at constant volume.
3. INTERNALCOMBUSTION ENGINE CYCLES
The internalcombustion engine utilizes as its working sub
stance a mixture of air and gas or air and petroleum vapor.
Combustion of the mixture takes place inside the engine cylinder,
or in a communicating vessel, and the heat generated is converted
into work. As the specific heat and the specific volume of the
mixture do not differ much from that of air, the cyclical analy
sis and the theory of the internalcombustion engine are developed
on the assumption that the working substance is air. The
internalcombustion engine cycles can be divided into the follow
ing groups:
Group I. Engines without compression.
Group II. Compression engines.
Group I. InternalCombustion Engines Without Compression
The Lenoir Engine. An engine of this group, invented by
Pierre Lenoir in i860, was the first successful practical gas
engine. The action of this type of engine is as follows :
As the piston leaves the dead center it draws into the cylinder
a charge of gas and air which is proportioned at the admission
54
CYCLES OF HEAT ENGINES USING GAS
valve to form an explosive mixture. Admission is cut off some
what before halfstroke and this is followed by the ignition of the
mixture by means of an electric spark. The explosion produces
a rapid rise in pressure above atmospheric, thus forcing the piston
to the end of the stroke. The exhaust valve opens near the end
of the stroke and the burnt products are expelled during the return
stroke of the piston. A flywheel carries the piston over during
the exhaust stroke as well as also, during the suction stroke.
The above series of operations takes place at each end of the
piston, producing two impulses for each revolution.
Fig. 10 represents the cycle of operations of the Lenoir engine
Volume
Fig. io. — Ideal Lenoir Cycle.
on PV coordinates. The drawing in of the mixture of gas and
air at atmospheric pressure is represented by ab. Ignition takes
place at b and is followed by the constant volume combustion
line be. The working stroke, an adiabatic expansion, takes place
to atmospheric pressure, as shown by cd, and the products of
combustion are rejected during the return stroke of the piston da.
Calling P, V, T the absolute pressure in pounds per square foot,
the specific volume in cubic feet per pound of mixture, and the
absolute temperature in degrees Fahrenheit respectively; also
using subscripts b, c, d to designate the points at the corners bed
of Fig. io, the following expressions will be obtained:
THE LENOIR ENGINE 55
If the heat developed by the complete combustion of the
mixture be designated by Q h then
6i = wC v (T c T b ). (79)
Since the combustion of the mixture takes place at constant
volume,
V c = V b
T c = T b + 2*. (80)
For unit weight,
p. = p> £• (81)
l 6
1
(82)
The expansion cd being adiabatic (n = 7),
If Q 2 is the heat rejected by the engine after performing its
cycle of operations,
Q 2 = wC v (T d  T b ). .(84)
The useful work available is :
W = J(Q 1  Q 2 ). (85)
The mean effective pressure is the average unbalanced pressure
on the piston of the engine in pounds per unit area, and using
symbols and units as on page 51,
M.E.P. = ir i V (S6)
V d — V b
The horse power developed would be found by equation
h.p. = K2S2L, (87)
33,000
where N designates the number of explosions per minute.
The cycle efficiency for the conditions of the problem is:
7? _ Qi ~ & ft _ T d — T b , RR x
E ^or I Q 1  t T zwr (88)
56
CYCLES OF HEAT ENGINES USING GAS
Group II. Compression Engines
The first compression engine was patented as early as 1799 by
Philip Lebon. The thermodynamic advantages of compression
before ignition will be evident from the demonstrations which
follow. A mechanical advantage due to compression is the re
duced shock due to the explosion and the resulting improved,
balance of parts.
1. The Otto Cycle. The Otto cycle, embodying the principles
first proposed by Beau de Rochas in 1862, resulted in the most
successful internalcombustion engine of today. This cycle was
adopted by Dr. Otto in 1876, the operations being carried out in
four consecutive equal strokes of the piston, requiring two com
plete revolutions of the engine crankshaft. The ideal diagram
Volume
Fig. 11. — Otto Cycle.
for the Otto cycle takes the form shown by Fig. 11, the operation
being as follows:
A mixture of gas and air is drawn in during the complete for
ward stroke of the piston, as shown by a' a. The return of the
piston compresses the mixture along the adiabatic curve ab.
Explosion of the compressed charge takes place at b, with the
consequent combustion at constant volume to c. cd is the adia
THE BRAYTON CYCLE 57
batic expansion producing the second forward stroke. The ex
haust valve opens at d, cooling the gases to the exhaust pressure a,
and rejecting them to the atmosphere.
The heat added during the combustion from b to c is
<2i = wC, (T c  T b ). (89)
The heat rejected from d to a is
ft = wC, (T d  T a ). (90)
The cycle efficiency is
E _ QiQ2 _ wC v {T c  T b ) wC, {T d  T a ) _ TgTg , .
Qx wC v {T c T b ) J T c T b ' l9Ij
Since the expansion and compression are adiabatic the follow
ing relations will hold:
T a Va n  1 = T b V b n ~\ and T c V c n  1 = TaV^ 1
d
Since V c = V b and V a = V d ,
d L c . I J d J a J a
7JT — 7fT> 2I S0 — — — — •
J a J b L c — J & i&.
Smce Tr(vj Ur
„ T a /VA" 1 /PA"— i \
E = I Tr I {va) =i {pJ n  (92)
Equation (92) shows that the efficiency of the Otto cycle de
pends on the amount of compression before explosion.
2. The Brayton Cycle. This cycle is often called the Joule
cycle after its inventor, or the Brayton cycle after George B.
Brayton, who in 1872 designed an engine with gradual constant
pressure combustion. In this engine a mixture of gas and air
is first compressed in a separate pump and forced into a receiver.
On the way from the receiver to the engine cylinder the mixture is
ignited by a gas jet which burns steadily without sudden expkx
sion, producing temperature and volume changes at constant
pressure. After expansion the piston drives out the products
of combustion at atmospheric pressure.
The cycle of operations for the Brayton engine is represented
58
CYCLES OF HEAT ENGINES USING GAS
by Fig. 12. a' a represents the supply of the combustible mixture
to the pump where it is compressed adiabatically to b and forced
into a receiver, be represents the burning of the compressed
6'
a'
b
c

VAdiabatic
\~~Adiabatic
^d
a
Volume
Fig. 12. — Bray ton Cycle.
mixture at constant pressure. As the mixture enters the working
cylinder, it expands adiabatically along cd. This is followed by
the rejection of the burnt gases along the atmospheric line da.
The heat added during the constant pressure combustion from
b to c is
Qi = wC P (Tc  T b ). (93)
The heat rejected from d to a is
Q 2 = wC P (T d — T a ). (94)
E
The cycle efficiency is
= Qi — Q2 = wCp (Tc — Tb) —wC p (Tg — Tg) _ Td — Tg , ,
Qi wC v (T c Tb) _I T c Tb {95)
Since the expansion and compression phases of the cycle are
adiabatic and P c = Pb\ Pa = Pa, therefore,
Td _ Tc
Tg Tb
J d La L a
TcTb ~ Y b " X
and
or =■©" «
Equation (96) is the same as equation (92) and shows that the
cycle efficiencies of the Otto and Br ay ton cycles are the same,
THE DIESEL CYCLE
59
under the same initial conditions, and depend on the amount of
compression of the charge before explosion.
3. The Diesel Cycle. This cycle is carried out in four strokes
of equal length, as in the case of the Otto cycle. Atmospheric
air is drawn in during the complete forward stroke of the piston
as shown by la (Fig. 13). The return of the piston compresses
Fig. 13. — Diesel Engine Cycle.
the air adiabatically to b, a pressure of about 500 pounds per
square inch. At the end of the compression stroke a charge of
the liquid fuel is injected in a finely divided form by an auxiliary
pump or compressor and burns nearly at constant pressure when
it comes in contact with the highly compressed air: The supply
and combustion of the fuel, as represented by the combustion
line be, is cut off at onetenth to onesixth the working stroke,
depending upon the load. Expansion takes place during the
balance of the stroke as shown by the adiabatic expansion line cd.
Release occurs at d with the consequent drop in pressure, and the
burnt gases are rejected during the fourth stroke of the cycle.
The heat added (Qi) during the combustion of the fuel is:
Q x = wC P (T c  T b ). (97)
The heat rejected from d to a is
Q 2 = wC v (T d  T a ). (98)
The cycle efficiency is then,
60 CYCLES OF HEAT ENGINES USING GAS
E „ ft Q2 „ wC P (Tc.  T b )  wC v (T d  T a )
Qi ' wC P (T c  T b )
C v (T d — T a \ I (T d — T a \ / x
" X C, Kf^Yj = 'yKf^Tj (99)
PROBLEMS
i. A Carnot engine containing 10 lbs. of air has at the beginning of the
expansion stroke a volume of 10 cu. ft. and a pressure of 200 lbs. per sq. in.
absolute. The exhaust temperature is o° F. If 10 B.t.u. of heat are added
to the cycle, find
(a) Efficiency of the cycle.
(b) Work of the cycle.
2. A Carnot cycle has at the beginning of the expansion stroke a pressure
of 75 lbs. per sq. in. absolute, a volume of 2 cu. ft. and a temperature of
200 F. The volume at the end of the isothermal expansion is 4 cu. ft.
The exhaust temperature is $0° F. Find
(a) Heat added to cycle.
(b) Efficiency of cycle.
(c) Work of cycle.
3. A cycle made up of two isothermal and two adiabatic curves has a
pressure of 100 lbs. per sq. in. absolute and a volume of 1 cu. ft. at the begin
ning of the isothermal expansion. At the end of the adiabatic expansion
the pressure is 10 lbs. per sq. in. absolute and the volume is 8 cu. ft. Find
(a) Efficiency of cycle.
(b) Heat added to cycle.
(c) Net work of cycle.
4. In a Carnot cycle the heat is added at a temperature of 400 F. and
rejected at 70 F. The working substance is 1 lb. of air which has a volume
of 2 cu. ft. at the beginning and a volume of 4 cu. ft. at the end of the
isothermal expansion. Find
(a) Volume at end of isothermal compression.
(b) Heat added to the cycle.
(c) Heat rejected from cycle.
(d) Net work of the cycle.
5. Air at a pressure of 100 lbs. per sq. in. absolute, having a volume of
1 cu. ft. and a temperature of 200 F., passes through the following opera
tions:
1st. Heat is supplied to the gas while expansion takes place under
constant pressure until the volume equals 2 cu. ft.
PROBLEMS 6l
2d. It then expands adiabatically to 15 lbs. per sq. in. absolute
pressure.
3d. Heat is then rejected while compression takes place at con
stant pressure.
4th. The gas is then compressed adiabatically to its original volume
of 1 cu. ft.
Find (a) Pounds of air used.
(b) Temperature at end of constant pressure expansion.
(c) Heat added to the cycle.
(d) Net work of cycle.
(e) Efficiency of cycle.
6. A Stirling hotair engine having a piston 16 inches in diameter and
a stroke of 4 ft. makes 28 r.p.m. The upper temperature is 650 F., while
the lower temperature is 150 F. Assuming that the volume of the working
cylinder is onehalf that of the displacer cylinder, calculate the pressures and
volumes at each point of the cycle (Fig. 8) ; also calculate the mean effective
pressure, the horse power developed, and the cycle efficiency.
7. Plot the Stirling engine cycle from the results of problem 6.
8. Compare the cycle efficiencies of internalcombustion engines working
on the Otto cycle using the fuels and compression pressures as indicated
below:
Gasoline, 75; producer gas, 150; and blast furnace gas, 200 lbs. per sq. in.
gage pressure.
9. Calculate the theoretical pressures, volumes and temperatures at each
point of an Otto cycle, as well as the mean effective pressure, horse power,
and cycle efficiency for the following conditions of Fig. 11. Assume that
the pressure after compression is 180 lbs. per sq. in. gage, that 80 B.t.u.
are added during combustion and that n in PV n = 1.4 :
P a = 14.7 lbs. per sq. in., V a = 13.5 cu. ft.,
T a = 7° + 4595 = 5 2 95° F. absolute.
10. Prove that cycle efficiency of the Diesel engine depends not only
upon the compression pressure before ignition but also upon the point of
cutoff C in Fig. 13.
11. In what respects do the actual cycles of internalcombustion engines
differ from the theoretical Otto and Diesel cycles?
12. The demonstrations in the text for the various gas cycles were based
on the assumption that the specific heats of gases are constant. Since the
specific heats of gases vary at high temperatures, show the effect of this
variability on the cyclic analysis. Refer to Lucke's Engineering Thermo
dynamics, Levin's Modern Gas Engines and the Gas Producer, and Clerk's
Gas, Petrol and Oil Engines, Vol. II.
CHAPTER V
PROPERTIES OF VAPORS
Saturated and Superheated Vapors. As was explained in the
chapter on the properties of perfect gases, a vapor can be liqui
fied by pressure or temperature changes alone. At every pres
sure there is a fixed point, called the point of vaporization, at
which a liquid can be changed into a vapor by the addition of
heat. A vapor near the point of vaporization is called a saturated
vapor and has a definite vaporization temperature for any given
pressure. When a vapor is heated so that its temperature is
greater than the vaporization temperature corresponding to a
given pressure, it is said to be a superheated vapor. Superheated
vapors only when far removed from the vaporization tempera
ture approach nearly the laws of perfect gases.
Theory of Vaporization. When heat is added to a liquid its
temperature will rise with a slight volume increase until the point
of vaporization is reached. This is always a definite point for
any given pressure and depends on the character of the liquid.
Thus the point of vaporization of water at the atmospheric pres
sure of 14.7 pounds per square inch is 21 2 F., while that at a pres
sure of 150 pounds absolute is 358 F. On the other hand, the
vaporization temperature of ammonia vapor at a pressure of 1 50
pounds absolute is 79 F. The increased temperature of vapori
zation with the pressure increase is due to the fact that the mole
cules being crowded, the velocity per molecule, or temperature,
must be great enough to overcome not only molecular attraction
but also external pressure before vaporization can take place.
When the point of vaporization is reached, any further addition
of heat will not cause any temperature increase, but internal
work accompanied by vaporization will be produced as well as
an enormous increase in volume. The heat required to entirely
62
VAPOR TABLES 63
vaporize a unit weight of a substance at a given pressure is termed
the latent heat of vaporization.
The volume developed when one unit weight of the liquid has
been completely evaporated is called the specific volume of the
dry vapor, this volume depending on the pressure.
When vaporization is incomplete the vapor is termed a wet
saturated vapor, which means that the vapor is in contact with
the liquid from which it is formed. The percentage dryness in
the vapor is called its quality. Thus the quality of steam is 0.97
when one pound of it consists of 97 per cent steam and 3 per cent
water. While the volume of a vapor increases with the quality,
the temperature is constant between the point of vaporization
and the point of superheat for any given pressure. Thus the
temperature of steam vapor corresponding to 150 pounds abso
lute is 358 F., no matter whether the quality is 0.10, 0.75 or 1.00.
Further addition of heat after complete vaporization will cause
temperature as well as volume changes, the substance being in
the superheated vapor condition.
Vapor Tables. The exact quantities of heat required to pro
duce the above effects under various conditions, as well as the
100
80
60
40
20
M =
Marl
1
s and Da%
is
G =
P =
Good
Peal)
enou
ody
ffh
M = 302.S
G = 302.9
P = 302.95
(
( M =
h=
292.7
292.7
292.7
Ui^.
<\
M = 267.3
G= 267.2 .
P = 267.26 (
IM = 281.0
'<G =281.0
i P=281.0
i^*""
'
M = 250.3 V
G =250.3
P=250.34
(
'M = 228.0
G = 228.0
P=327.95
f M =
i G p =
212°
212°
211.9
!6
\
240 260 280
Temperature, Degrees Fahrenheit
220
Fig. 14. — PressureTemperature Relations for Saturated Steam
300
relations existing between pressure, volume, and temperature of
saturated and superheated vapors have been determined exper
6 4
PROPERTIES OF VAPORS
imentally. These experimental results have been given in the
form of empirical equations from which vapor tables have been
computed. Tables 3 and 4, showing the properties of dry satu
100
•3 80
TIX
<
a
• 60
a*
40
20
12 14 16 18
Volume, cu. ft. per lb.
Fig. 15. — Pressure Volume Relations for Saturated Steam.
M = Mark
G = Good
P = Peat
s and Da 1
enough
ody
as
P^ '
880
Fig. 16.
900 920 940 960
Latent Heat
PressureLatent Heat Relations for Saturated Steam.
rated steam and of ammonia, are given in the appendix. This
text should be supplemented by the more complete tables of
Marks and Davis, Peabody or Goodenough.
TEMPERATURE, PRESSURE AND VOLUME OF STEAM 65
In Figs. 14, 15 and 16 are plotted the several variables, as given
in the abovementioned tables, for various pressures.
In most vapor tables will be found, corresponding to the pres
sure of the vapor in pounds per square inch absolute (p), the
vaporization temperature (t), the heat of the liquid (q, h or i'),
the heat of vaporization (r or L), the specific volume (s or v),
entropy of water (0 , 5 or n), entropy of vaporization ( — or — )>
density pounds per cubic foot I or J, internal latent heat
(p, i or I). In some tables will also be found a column for the
total heat of vapor (H = q + r), external latent heat (APu)
and entropy.
Relation between Temperature, Pressure and Volume of
Saturated Steam. The important relations of temperature, pres
sure and volume were first determined in a remarkable series of
experiments conducted by a French engineer named Regnault,
and it has been on the basis of his data, first published in 1847,
that even our most modern steam tables were computed. Later
experimenters have found, however, that these data were some
what in error, especially for values near the dry saturated
condition. These errors resulted because it was difficult in the
original apparatus to obtain steam entirely free from moisture.
The pressure of saturated steam increases very rapidly as the
temperature increases in the upper limits of the temperature scale.
It is very interesting to examine a table of the properties of
steam to observe how much more rapidly the pressure must be
increased in the higher limits for a given range of temperature.
It should be observed that in most tables the pressure is almost
invariably given in terms of pounds per square inch, while in
nearly ail our thermodynamic calculations the pressure must
be used in pounds per square foot.
The specific volume of saturated steam (v or s) is equal to
the sum of the volumes of water (a) and of the increase in
volume during vaporization («) , or
V = a + U. (100)
66 PROPERTIES OF VAPORS
Heat in the Liquid (Water) (h or q). The essentials of the
process of making steam have been described in a general way.
The relation of this process to the amount of heat required will
now be explained. If a pound of water which is initially at
some temperature t is heated at a constant pressure P (pounds
per square foot) to the boiling point corresponding to this pres
sure and then converted into steam, heat will first be absorbed
in raising the . temperature of the water from to to t, and then
in producing vaporization. During the first stage, while the
temperature is rising, the amount of heat taken in is approxi
mately (t — t ) heat units, that is, British thermal units (B.t.u.),
because the specific heat of water is approximately unity and
practically constant. This number of B.t.u. multiplied by 778
gives the equivalent number of footpounds of work. For the
purpose of stating in steam tables the amount of heat required for
this heating of water, the initial temperature t must be taken at
some definite value; for convenience in numerical calculations
and also because of long usage, the temperature 32 F. is invari
ably used as an arbitrary starting point for calculating the
amount of heat " taken in." The symbol h (or sometimes i' or q)
is used to designate the heat required to raise one pound of
water from 32 F. to the temperature at which it is vaporized into
steam. In other words, " the heat of the liquid" (h) is the
amount of heat in B.t.u. required to raise one pound of water
from 3 2 F. to the boiling point, or:
h = \C dt. (101)
C is the specific heat of water at constant pressure, U is the
freezing point of water, / is the temperature to which the water
is raised.
As C is very nearly unity, the value of the heat absorbed by
water in being raised to the steaming temperature (h) in B.t.u.
can be expressed, approximately, by the formula
h = t — 32 (in B.t.u.). (102)
Values of h, taking into consideration the variation in the
EXTERNAL WORK OF EVAPORATION 67
specific heat of water, will be found in the usual steam tables
(Table 3).
During this first stage, before any steaming has occurred, prac
tically all the heat applied is used to increase the stock of internal
energy. The amount of external work done by the expansion
of water as a liquid is practically negligible.
Latent Heat of Evaporation (L or r). In the second stage of the
formation of steam as described, the water at the temperature t,
corresponding to the pressure, is changed into steam at that
temperature. Although there is no rise in temperature, very
much heat is nevertheless required to produce this evaporation
or vaporization. The heat taken in during this stage is the
latent heat of steam. In other words, the latent heat of steam
may be defined as the amount of heat which is taken in by a
pound of water while it is changed into steam at constant pres
sure, the water having been previously heated up to the tem
perature at which steam forms. The symbol L (also sometimes
r) is used to designate this latent heat of steam. Its value
varies with the particular pressure at which steaming occurs
being somewhat smaller in value at high pressures than at low.
The latent heat (L or r) of saturated steam can be approxi
mately calculated by the formula:
L = 97O.4 — O.655 Q — 212) — O.OOO45 (t — 2T2) 2 . (103)
External Work of Evaporation (Apu). A part of the heat taken
in during the " steaming " process is spent in doing external
work. Only a small part of the heat taken in is represented
by the external work done in making the steam in the boiler,
and the remainder of the latent heat (L) goes to increase the
internal energy of the steam. The amount of heat that goes
into the performing of external work is equal to P (the pressure
in pounds per square foot) times the change of volume (u)
occurring when the water is changed into steam, or Pu.
Example. At the usual temperatures of the working fluid in
steam engines the volume of a pound of water a is about sV of a
cubic foot. The external work done in making one . pound of
68 PROPERTIES OF VAPORS
steam having finally a volume of V (cubic feet) at a constant pres
sure P (pounds per square foot) may be written in footpounds:
External work = Pu = P (V — gV). C 1 ^)
This last equation can be expressed in British thermal units
(B.t.u.) by APu where A = yfg. It is apparent also from this
equation that the external work done in making steam is less at
low pressure than at high,* because there is less resistance to over
come, or, in other words, P in equation (104) is less. The heat
equivalent of the external work is, therefore, a smaller propor
tion of the heat added at low temperature than at high.
Total Heat of Steam (H). The heat added during the process
represented by the first and second stages in the formation of a
pound of steam, as already described, is called the total heat of
saturated steam or, for short, total heat of steam, and is repre
sented by the symbol H. Using the symbols already defined,
we can write, per pound of steam,
H = h + L (B.t.u.). (105)
In other words, this total heat of steam is the amount of heat
required to raise one pound of water from 32 F. to the tempera
ture of vaporization and to vaporize it into dry steam at that
temperature under a constant pressure.
Remembering that h for water is approximately equal to the
temperature less 32 degrees corresponding to the pressure at
which the steam is formed (t) the total heat of steam is approxi
mately,
H = (t  32) + L. (106)
To illustrate the application of equation (106) calculate the
total heat of steam being formed in a boiler at an absolute pres
sure of 1 1 5 pounds per square inch.
* Although at the lower pressure the volume of a given weight of steam is
greater than at a higher pressure, the change of pressure is relatively so much greater
in the process of steam formation that the product of pressure and change of
volume, P (V2 — Vi), which represents the external work done, is less for low
pressure steam than for high.
INTERNAL ENERGY OF EVAPORATION AND OF STEAM 69
From the steam tables (Table 3) the temperature t of the
steam at a pressure of 115 pounds per square inch absolute is
338 F., the latent heat of vaporization is 880 B.t.u. per pound
and the total heat of the steam is 1189 B.t.u. per pound. To
check these values with equation (106), by substituting values of
L and t,
E = (338  3 2 ) + 880 = 1186 B.t.u. per pound.
When steam is condensed under constant pressure, the process
which was called the " second stage " is reversed and the amount
of heat equal to the latent heat of evaporation (L) is given up
during the change that occurs in the transformation from steam
to water.
Internal Energy of Evaporation and of Steam. It was ex
plained in a preceding paragraph that when steam is forming
not all of the heat added goes into the internal or " intrinsic "
energy of the steam, but that a part of it was spent in performing
external work. If, then, the internal energy of evaporation is
represented by the symbol I L ,
ULPZ^. (107)
This equation represents the increase in internal energy which
takes place in the changing of a pound of water at the tem
perature t into steam at the same temperature. In all the
formulas dealing with steam the state of water at 32 F. has been
adopted as the arbitrary starting point from which the taking in
of heat was calculated. This same arbitrary starting point is
used also in expressing the amount of internal energy in the steam.
This is the excess of the heat taken in over the external work done
in the process.
The total internal energy (L H ) of a pound of saturated steam at
a pressure P in pounds per square foot is equal to the total heat
(H) less the heat equivalent of the external work done; thus,
i a = np {V ~/ t)  (108)
778
70 PROPERTIES OF VAPORS
Such reference is made here to the internal energy of steam
because it is very useful in calculating the heat taken in and
rejected by steam during any stage of its expansion or com
pression.
Heat taken in = increase of internal energy + external work
done.
When dealing with a compression instead of an expansion the
last term above (external work) will be a negative value to indi
cate that work is done upon the steam instead of the steam doing
work by expansion.
The following problem shows the calculation of internal energy
and external work:
Example. A boiler is evaporating water into dry and satu
rated steam at a pressure of 300 pounds per square inch abso
lute. The feed Water enters the boiler at a temperature of 145 F.
The internal energy of evaporation per pound of steam is
I  l P ( V ~ ^ = Sn 3 °° X I44 ^' 551 ~ 7 ^
778 " ' 6 778 
= 811.3 — 85.3 = 726.0 B.t.u.,
or taken directly from saturated steam tables equals 726.8 B.t.u.
The total internal energy supplied above 32 F. per pound of
steam is
I H = H  P ( V ~ ^ = 1204.1  85.3 = 1118.8 B.t.u.,
or taken directly from saturated steam tables equals 11 18.5 B.t.u.
External work done above 32 F. per pound of steam as calcu
lated from steam tables is
H — I H = 1204.1 — 1118.5 = 85.6 B.t.u.
External work of evaporation per pound of steam as calcu
lated from the steam tables is
L I L = 811.3  726.8 = 84.5 B.t.u.
The external work done in raising the temperature of a pound
of water from 32 F. to the boiling point is
85.6  84.5 = 1.1 B.t.u.
WET STEAM 7 1
The external work done in raising the temperature of a pound
of water from 3 2° F. to 145 F. is
300 X 144 (0.0163 — 0.01602) p.
£ ttfL ^  = 0.01 B.t.u.
778
The external work done in forming the steam from water at
145 F. is, then,
84.5 + 1. 10 — 0.01 = 85.59 B.t.u.
as compared with 85.3 B.t.u. as calculated from
P (V ~ A)
778
It is to be noticed that the external work done during the addi
tion of heat when the liquid is raised to the vaporization tem
perature is small as compared with other values, and for most
engineering work it is customary to assume that no external work
is done during the addition of heat to the water. With this
assumption
I H = h + L Piy ~ A) =H  (Z,  I L ). (109)
778
Steam Formed at Constant Volume. When saturated steam
is made in a boiler at constant volume, as, for example, when the
piping connections from the boiler to the engines are closed, then
no external work is done, and all the heat taken in is converted
into and appears as internal energy I H of the steam. This
quantity is less than the total heat H of steam, representing its
formation at constant pressure, by quantity P (V — ^V) 5 778,
where P represents the absolute pressure at which the steam is
formed in pounds per square foot and V is the volume of a
pound of steam at this pressure in cubic feet.
Wet Steam. In all problems studied thus far, dealing with
saturated steam, it has been assumed that the steaming process
was complete and that the water had been completely converted
into steam. In actual practice it is not at all unusual to have
steam leaving boilers which is not perfectly and completely vapor
72 PROPERTIES OF VAPORS
ized; in other words, the boilers are supplying to the engines a
mixture of steam and water. This mixture we call wet steam.
It is steam which carries actually in suspension minute particles
of water, which remain thus in suspension almost indefinitely.
The temperature of wet steam is always the same as that of dry
saturated steam as given in the steam tables, so long as any
steam remains uncondensed.
The ratio of the weight of moisture or water in a pound of wet
steam to a pound of completely saturated steam is called the
degree of wetness; and when this ratio is expressed as a per
cent, it is called the percentage of moisture or " per cent wet."
If in a pound of wet steam there is 0.04 pound of water in suspen
sion or entrained, the steam is four per cent wet.
Another term, called the quality of steam, which is usually
expressed by the symbol x, is also frequently used to represent
the condition of wet steam. Quality of steam may be denned
as the proportion of the amount of dry or completely evaporated
steam in a pound of wet steam. To illustrate with the example
above, if there is 0.04 pound of water in a pound of wet steam;
the quality in this case would be 1 — 0.04 or 0.96. In this case
the quality of this steam is ninetysix onehundredths.
With this understanding of the nature of wet steam it is obvi
ous that latent heat of a pound of wet steam is xL. Similarly,
the total heat of a pound of wet steam is h + xL, and the
volume of a pound of wet steam is x (V — ^V) + ih — xV + ^
(1 — x), or approximately it is equal to xV, because the term A
(1 — x) is negligibly small except in cases where the steam is so
wet as to consist mostly of water. Similarly, the internal energy
in a pound of wet steam is
Superheated Steam. When the temperature of steam is higher
than that corresponding to saturation as taken from the steam
tables, and is, therefore, higher than the standard temperature
corresponding to the pressure, the steam is said to be super
THE TOTAL HEAT OF SUPERHEATED STEAM 73
heated. In this condition steam begins to depart and differ
from its properties in the saturated condition, and when super
heated to a very high degree it begins to behave somewhat
like a perfect gas.
There are tables of the properties of superheated steam just
as there are tables of saturated steam. (See Marks and Davis',
Goodenough's or Peabody's Steam Tables.) When dealing with
saturated steam there is always only one possible temperature and
only one specific volume to be considered. With superheated
steam, on the other hand, for a given pressure there may be any
temperature above that of saturated steam, and corresponding to
each temperature there will be, of course, definite values for
specific volume and total heat. Like a perfect gas the specific
volume or the cubic feet per pound increases with the increase in
temperature.
The total heat of superheated steam is obviously greater than
the total heat of saturated steam.* Thus, since the total heat
of dry saturated steam is, as before, h + L, the total heat of
superheated steam with D degrees of superheat is
Us = h + L + C p X D; (in)
or if the temperature of the superheated steam is / sup . and / sa t. is
the temperature of saturated steam corresponding to the pressure,
H s = h + L + C p (4u P . — feat.). (112)
H s (equations in and 112) is the total heat of superheated
steam or is the amount of heat required to produce a pound of
steam with the required degrees of superheat from water at 32 F.
To obtain the amount of internal energy of a pound of steam
(superheated) corresponding to this total heat as stated above,
the external work expended in the steaming process must be
subtracted; that is,
* In practice steam is called dry saturated when it is exactly saturated and has
no moisture. It is the condition known simply as saturated steam as regards the
properties given in the ordinary steam tables. The dry saturated condition is the
boundary between wet steam and superheated steam.
74 PROPERTIES OF VAPORS
I H = h + L  P (F Sat  ~ A) + Cp (/ sup ,  feaO  jP(7sUP ~ Fsat  )
778 778
= ft  1; — Lsp \tsup. fsat.j r
7
778 ^ 0/
The term ^ can be neglected in the equations above for prac
tically all engineering calculations as the maximum error from
this is not likely to be more than one in one thousand or xV per
cent.* The accuracy of our steam tables for values of latent and
total heat is not established to any greater degree. Making
this approximation, the equation above becomes
Ih = h f L f Cp (/sup. — /sat.) — ( • ) = H s ''
\ 778 / 778
If examination is made of the properties of superheated steam
as given in Marks and Davis' or in Goodenough's Steam Tables
for superheated steam at 165 pounds per square inch absolute
pressure and 150 F. superheat, the following results are secured:
Marks & Davis Goodenough
Temperature (/) 516.0 516.1
Volume (7, v) 3.43 34i
Total Heat (H, or h\ or i) 1277.6 1280.8
The values of h and L for saturated steam at this pressure are
respectively 338.2 and 856.8. From the curves given, Fig. i7,{
* The error in the value of total internal energy due to neglecting the term fa
in the exercise worked out on page 70 is 0.85 B.t.u. per pound (one per cent of the
external work), or an error of .076 per cent in the final result.
f In their tables Marks and Davis represent the total heat of superheated
steam by h. In this book the symbol Hs is used for greater clearness and to be
consistent with the symbols used in the preceding formulas. They use also v for
specific volume in place of V as above.
J The curves for values of Cp, as given in Fig. 17, are for average and not for
instantaneous values such as are given in Fig. 18, of Marks and Davis' Tables and
Diagrams. Great caution must be observed in the use of curves of this kind.
Those giving instantaneous values can only be used for the value of Cp in the form
ulas given for the total heat of superheated steam after the average value has been
found by integrating the curve representing these values for a given pressure.
DRYING OF STEAM BY THROTTLING OR WIREDRAWING 75
the specific heat of superheated steam at constant pressure (C p )
is 0.552. From these data,
E s = h + L + C v X 150 = 338.2 + 856.8 + 0.552 X 150 = 1277.7.
This value agrees very nearly with that given in the tables as
previously indicated.
.66
'J4
M
.SO
■m
* a J»
"g oq .56
S 
 1 .54
Lb
s. ,
Vbs
 227.2
"
A 0, .52
3 ?
"170.4
142.2
113.6
4!
a ^0
 85.2
»
I

5C.8
»
48

1 ■
— 28.4
,,
[
■*  "
.46
,.44
,.42
Fig. 17.
200 250 300 350 400 450 500 550 COO 650 700 750
Temperature ° P
 Mean Values of Cp Calculated by Integration from Knoblauch
and Jakob's Data.
V = \ 0.5962 T—p (1+0.0014^) (
■0.0833
> (114)
IP
The specific volume (V) is calculated from the empirical form
ula derived from experimental results and is expressed as follows:
'150 ,300,000
where p is in pounds per square inch, V is in cubic feet per pound
and T = t + 460 is the absolute temperature on the Fahrenheit
scale.*
Drying of Steam by Throttling or Wiredrawing. When steam
expands by passing through a very small opening, as, for example,
* The value of 7 or ■= of superheated steam used in ordinary engineering
calculations is 1.3.
7 6
PROPERTIES OF VAPORS
through a valve only partly open in a steam line, the pressure
is considerably reduced. This effect is called throttling or wire
drawing. The result of expansion of this kind when the pressure
O.90
ioo°
150
200 250 o 300
temperature G.
•350
Fig. i8. — Values of the "True" Specific Heat of Superheated Steam.
is reduced and no work is done is that if the steam is initially
wet it will be drier and if it is initially dry or superheated the
degree of superheat will be increased. The reason for this is
that the total heat required to form a pound of dry saturated
steam (H) is considerably less at low pressure than at high, but
obviously the total quantity of heat in a pound of steam must be
THROTTLING OR SUPERHEATING CALORIMETERS 77
the same after wire drawing as it was before, neglecting radiation.
Now if steam is initially wet and the quality is represented by
Xi, then the total heat in the steam is represented by fa plus XiLi,
in which fa and Li represent the heat of the liquid and the latent
heat of the steam at the initial pressure. If, also, the quality,
heat of liquid and the latent heat of the steam after wiredrawing
are represented respectively by X2, fa and Z2, then fa .+ X1L1
= fa + X2L2, or
X1L1 + fa  fa . N
X2=  (115)
This drying action of steam in passing through a small open
ing or an orifice is very well illustrated by steam discharging
from a small leak in a high pressure boiler into the atmosphere.
It will be observed that no moisture is visible in the steam a
few inches from the leak but farther off it becomes condensed
by loss of heat, becomes clouded and plainly visible. An im
portant application of the throttling principle is also to be found
in the throttling calorimeter (page 80).
Determination of the Moisture in Steam. Unless the steam
used in the power plant is superheated it is said to be either dry
or wet, depending on whether or not it contains water in sus
pension. The general types of steam calorimeters used to de
termine the amount of moisture in the steam may be classified
under three heads:
1 . Throttling or superheating calorimeters.
2. Separating calorimeters.
3. Condensing calorimeters.
Throttling or Superheating Calorimeters. The type of steam
calorimeter used most in engineering practice operates by passing
a sample of the steam through a very small orifice, in which it
is superheated by throttling. A very satisfactory calorimeter of
this kind can be made of pipe fittings as illustrated in Fig. 19.
It consists of an orifice O discharging into a chamber C, into
which a thermometer T is inserted, and a mercury manometer
78 PROPERTIES OF VAPORS
is usually attached to the cock V 3 for observing the pressure in
the calorimeter.
It is most important that all parts of calorimeters of this
type, as well as the connections leading to the main steam pipe,
should be very thoroughly lagged by a covering of good insu
lating material. One of the best materials for this use is hair
felt, and it is particularly well suited for covering the more or
less temporary pipe fittings, valves and nipples through which
steam is brought to the calorimeter. Throttling calorimeters
have been found useless because the small pipes leading to the
calorimeters were not properly lagged, so that there was too
much radiation, producing, of course, condensation, and the
calorimeter did not get a true sample. It is obvious that if the
entering steam contains too much moisture the drying action
due to the throttling in the orifice may not be sufficient to super
heat. It may be stated in general that unless there are about
5° to io° F. of superheat in the calorimeter, or, in other words,
unless the temperature on the low pressure side of the orifice is
at least about 5 to io° F. higher than that corresponding to the
pressure in the calorimeter, there may be some doubt as to the
accuracy of results.* The working limits of throttling calo
rimeters vary with the initial pressure of the steam. For 35
pounds per square inch absolute pressure the calorimeter ceases
to superheat when the percentage of moisture exceeds about
2 per cent; for 150 pounds absolute pressure when the moisture
exceeds about 5 per cent; and for 250 pounds absolute pressure
when it is in excess of about 7 per cent. For any given pressure
the exact limit varies slightly, however, with the pressure in the
calorimeter.
* The same general statement may be made as regards determinations of
superheat in engine and turbine tests. Experience has shown that tests made
with from o to 10 degrees Fahrenheit superheat are not reliable, and that the
steam consumption in many cases is not consistent when compared with results
obtained with wet or more highly superheated steam. The errors mentioned
when they occur are probably due to the fact that in steam, indicating less than
10 degrees Fahrenheit superheat, water in the liquid state may be taken up in
"slugs" and carried along without being entirely evaporated.
THROTTLING OR SUPERHEATING CALORIMETERS 79
In connection with a report on the standardizing of engine
tests, the American Society of Mechanical Engineers * published
instructions regarding the method to be used for obtaining a
fair sample of the steam from the main pipes. It is recom
mended in this report that the calorimeter shall be con
nected with as short intermediate piping as possible with a
socalled calorimeter nipple made of Jinch pipe and long enough
to extend into the steam pipe to within  inch of the opposite
wall. The end of this nipple is to be plugged so that the steam
must enter through not less than twenty finch holes drilled
around and along its length. None of these holes shall be less
than J inch from the inner side of the steam pipe. The sample
of steam should always be taken from a vertical pipe as near as
possible to the engine, turbine or boiler being tested. A good
example of a calorimeter nipple is illustrated in Fig. 20.
The discharge valve V 2 should not be closed or adjusted
without' first closing the gage cock V 3 . Unless this precaution is
taken, the pressure may be suddenly increased in chamber C, so
that if a manometer is used the mercury will be blown out of it;
and if, on the other hand, a lowpressure steam gage is used it
may be ruined by exposing it to a pressure much beyond its scale.
Usually it is a safe rule to begin to take observations of tem
perature in calorimeters after the thermometer has indicated a
maximum value and has again slightly receded from it. The
quality or relative dryness of wet steam is easily calculated by
the following method and symbols :
pi = steam pressure in main, pounds per square inch
absolute,
p 2 = steam pressure in calorimeter, pounds per square
inch absolute,
t c = temperature in calorimeter, degrees Fahrenheit,
L\ and hi = heat of vaporization and heat of liquid corre
sponding to pressure p h B.t.u.,
H 2 and t 2 = total heat (B.t.u.) and temperature (degrees
Fahrenheit) corresponding to pressure p 2 ,
* Proceedings American Society of Mechanical Engineers, vol. XXI.
8o
PROPERTIES OF VAPORS
Op
specific heat of superheated steam. Assume 0.47
for low pressures existing in calorimeters,
initial quality of steam.
Connection for
Manometer
Fig. 19. — Simple Throttling Steam Calorimeter.
Total heat in a pound of wet steam flowing into orifice is
XiU + h h
and after expansion, assuming all the moisture is evaporated,
the total heat of the same weight of steam is
H 2 ~r Lp \pc — W).
Then assuming no heat losses and putting for C p its value 0.47 :
X1L1 + h = H 2 + 0.47 (t e — t 2 ) (116)
H 2 + 0.47 (t c — t 2 ) — hi
Xi =
JU
(117)
BARRUS, THROTTLING CALORIMETER 8l
The following example shows the calculations for finding the
quality of steam from the observations taken with a throttling
calorimeter :
Example. Steam at a pressure of ioo pounds per square inch
absolute passes through a throttling calorimeter. In the calo
rimeter the temperature of the steam becomes 243 ° F. and the
pressure 15 pounds per square inch absolute. Find the quality.
Solution. By taking values directly from tables of properties
of superheated steam,* the total heat of the steam in the calorim
eter at 15 pounds per square inch absolute pressure and 243 ° F.
is 1 164.8 B.t.u. per pound or can be calculated as follows:
E + C v {t c  t 2 ) = 1150.7 + 0.47 ( 2 43  2I 3)
= 1 164.8 B.t.u. per pound.
(Note. t c = temperature in calorimeter and t 2 = temperature
corresponding to calorimeter pressure.)
The total heat of the steam before entering the calorimeter
is h + xL. At 100 pounds per square inch absolute pressure,
this is 298.3 + 888.0 x in B.t.u. per pound. Since the heat in
the steam per pound in the calorimeter is obviously the same as
before it entered the instrument, we can equate as follows:
298.3 + 888.0 x = 1164.8
x = 0.976
or the steam is 2.4 per cent wet.
Barrus Throttling Calorimeter. This is an important varia
tion from the type of throttling calorimeter shown in Fig. 19 and
has been quite widely introduced by Mr. George H. Barrus. In
this apparatus the temperature of the steam admitted to the
calorimeter is observed instead of the pressure and a very free
exhaust is provided so that the pressure in the calorimeter is
atmospheric. This arrangement simplifies very much the obser
vations to be taken, as the quality of the steam X\ can be calcu
lated by equation (117) by observing only the two temperatures
* Marks and Davis' Steam Tables and Diagrams (1st ed.), page 24 or
Goodenough's Properties of Steam and Ammonia (2d ed.), page 47.
82
PROPERTIES OF VAPORS
Fig. 20.
Barms' Throttling Steam
Calorimeter.
/1 and t c , taken respectively on the high and low pressure sides of
the orifice in the calorimeter. This calorimeter is illustrated in
Fig. 20. The two thermometers required are shown in the
figure. Arrows indicate the path of the steam.*
The orifice in such calorimeters is usually made about ft inch
in diameter, and for this size of orifice the weight of steam 
discharged per hour at 175 pounds
per square inch absolute pressure
is about 60 pounds. It is impor
tant that the orifice should always
be kept clean, because if it be
comes obstructed there will be a
reduced quantity of steam passing
through the instrument, making
the error due to radiation rela
tively more important.
In order to free the orifice from
dirt or other obstructions the connecting pipe to be used for
attaching the calorimeter to the main steam pipe should be
blown out thoroughly with steam before the calorimeter is put
in place. The connecting pipe and valve should be covered
with hair felting not less than f inch thick. It is desirable also
that there should be no leak at any point about the apparatus,
as in the stuffingbox of the supply valve, the pipe joints or the
union.
With the help of a diagramf giving the quality of steam directly
the Barrus calorimeter is particularly well suited for use in
power plants, where the quality of the steam is entered regu
larly on the log sheets. The percentage of moisture is obtained
immediately from two observations without any calculations.
Separating Calorimeters. It was explained that throttling
calorimeters cannot be used for the determination of the quality
of steam when for comparatively low pressures the moisture
* Transactions of American Society of Mechanical Engineers, vol. XI, page 790.
f In boiler tests corrections should be made for the steam discharged from the
steam calorimeters.
J See page 113, and also Moyer's Power Plant Testing, (2d ed.), pages 5860.
SEPARATING CALORIMETERS
83
z = Steam
Water
is in excess of 2 per cent, and when for boiler pressures commonly
used it exceeds 5 per cent. For higher percentages of moisture
than these low limits separating calorimeters are most generally
used. In these instruments the water is removed from the
sample of steam by mechanical separation just as it is done
in the ordinary steam separator installed in the steam mains
of a power plant. There is pro
vided, of 'course, a device for de
termining, while the calorimeter is
in operation, usually by means of
a calibrated gage glass, the amount
of moisture collected. This me
chanical separation depends for its
action on changing very abruptly
the direction of flow and reducing
the velocity of the wet steam.
Then, since the moisture (water)
is nearly 300 times as heavy as
steam at the usual pressures de
livered to the engine, the moisture
will be deposited because of its
greater inertia.
Fig. 21 illustrates a form of sep
arating calorimeter having a steam
jacketing spaqe which receives live
steam at the same temperature as the sample. Steam is supplied
through a pipe A, discharging into a cup B. Here the direction
of the flow is changed through nearly 180 degrees, causing the
moisture to be thrown outward through the meshes in the cup
into the vessel V. The dry steam passes upward through the
spaces between the webs W, into the top of the outside jacketing
chamber J, and is finally discharged from the bottom of this
steam jacket through the nozzle N. This nozzle is consider
ably smaller than any other section through which the steam
flows, so that there is no appreciable difference between the pres
sures in the calorimeter proper and the jacket. The scale
Fig. 21. — Separating Calorimeter.
84 PROPERTIES OF VAPORS
opposite the gage glass G is graduated to show, in hundredths
of a pound, at the temperature corresponding to steam at
ordinary working pressures, the variation of the level of the
water accumulating. A steam pressure gage P indicates the
pressure in the jacket J, and since the flow of steam through
the nozzle N is roughly proportional to the pressure, another
scale in addition to the one reading pressures is provided
at the outer edge of the dial. A petcock C is used for draining
the water from the instrument, and by weighing the water col
lected corresponding to a given difference in the level in the gage
G the graduated scale can be readily calibrated. Too much
reliance should not be placed on the readings for the flow of steam
as indicated by the gage P unless it is frequently calibrated.
Usually it is very little trouble to connect a tube to the nozzle N
and condense the steam discharged in a large pail nearly filled
with water. When a test for quality is to be made by this method
the pail nearly filled with cold water is carefully weighed, and then
at the moment when the level of the water in the water gage G
has been observed the tube attached to the nozzle N is immedi
ately placed under the surface of the water in the pail. The
test should be stopped before the water gets so hot that some
weight is lost by " steaming." The gage P is generally cali
brated to read pounds of steam flowing in ten minutes. For the
best accuracy it is desirable to use a pail with a tightly fitting
cover into which a hole just the size of the tube has been cut.
If W is the weight of dry steam flowing through the orifice N
and w is the weight of moisture separated, the quality of the
steam is
W + w
Condensing or Barrel Calorimeter. For steam having a large
percentage of moisture (over 5 per cent) the condensing or
barrel calorimeter will give fairly good results if properly used.
In its simplest form it consists of a barrel placed on a platform
scale and containing a known weight of cold water. The steam
is introduced by a pipe reaching nearly to the bottom of the
EQUIVALENT EVAPORATION AND FACTOR OF EVAPORATION 85
barrel. The condensation of the steam raises the temperature
of the water, the loss of heat by the wet steam being equal to
the gain of heat by the cold water. It will, therefore, be neces
sary to observe the initial and final weights, the initial and
final temperatures of the water in the barrel and the temperature
of the steam.
Let W = original weight of cold water, pounds.
w = weight of wet steam introduced, pounds.
tx — temperature of cold water, degrees Fahrenheit.
t 2 = temperature of water after introducing steam.
t s = temperature of steam.
L = latent heat of steam at temperature t s .
x = quality of steam.
Then,
Heat lost by wet steam = heat gained by water.
wxL + w (t s — h) = W (t 2 — h)
X = wl (lI9)
The accuracy of the results depends obviously upon the accu
racy of the observation, upon thorough stirring of the water so
that a uniform temperature is obtained and upon the length of
time required. The time should be just long enough to obtain
accurate differences in weights and temperatures; otherwise,
losses by radiation will make the results much too low.
Equivalent Evaporation and Factor of Evaporation. For the
comparison of the total amounts of heat used for generating
steam (saturated or superheated) under unlike conditions it. is
necessary to take into account the temperature t Q at which the
water is put into the boiler as well as also the pressure P at
which the steam is formed.* These data are of much impor
tance in comparing the results of steam boiler tests. The basis
* As the pressure P increases the total heat of the steam also increases; but
as the initial temperature of the water ("feed temperature") increases the value
of the heat of the liquid decreases.
86 PROPERTIES OF VAPORS
of this comparison is the condition of water initially at the boil
ing point for "atmospheric" pressure or at 14.7 pounds per
square inch; that is, at 212 F. and with steaming taking place
at the same temperature. For this standard condition, then,
h = o and H = L = 970.4 B.t.u. per pound. Evaporation
under these conditions is described as,
" from (a feedwater temperature of) and at (a pressure corre
sponding to the temperature of) 212 F."
To illustrate the application of a comparison with this stand
ard condition, let it be required to compare it with the amount
of heat required to generate steam at a pressure of 200 pounds
per square inch absolute with the temperature of the water sup
plied (feed water) at 190 F.
For P = 200 pounds per square inch absolute the heat of the
liquid h = 354.9 B.t.u. per pound and the heat of evaporation
(L) is 843.2 B.t.u. per pound. For t = 190 F. the heat of the
liquid (ho) is 157.9 B.t.u. per pound. The total heat actually
required in generating steam at these conditions is, therefore,
843.2 + (354.9 — I579) = 10402 B.t.u. per pound.
The ratio of the total heat actually used for evaporation to
that necessary for the condition defined by " from and at 212 F."
is called the factor of evaporation. In this case it is the value
1040.2 t 970.4 = 1.07.
Calling F the factor of evaporation, h and L respectively the
heats of the liquid and of evaporation corresponding to the steam
pressure and h Q the heat of the liquid corresponding to the tem
perature of feed water, then
^ = L + (hho) . (i2q)
970.4
The actual evaporation of a boiler (expressed usually in pounds
of steam per hour) multiplied by the factor of evaporation is
called the equivalent evaporation.
PROBLEMS 87
Vapors as Refrigerating Media. Ammonia is generally used
as the refrigerating medium in connection with mechanical
refrigeration. Carbon dioxide and sulphur dioxide are also used
to a limited extent as refrigerating media.
The value of a substance as a refrigerating medium depends
upon its latent heat, its vaporization temperature, its cost, and
upon its chemical properties. The greater the latent heat of a
refrigerating medium the more heat will it be capable of abstract
ing by evaporation. Upon the vaporization temperature of the
refrigerant at different pressures depends the degree of cold it
can produce as well as its practicability for use in hot climates
where low temperature cooling water cannot be secured.
Goodenough's Tables of Properties of Steam and Ammonia give
the various physical properties of saturated, superheated and
liquid ammonia. Peabody's Tables give the properties of
ammonia, sulphur dioxide, carbon dioxide and of other vapors
used as refrigerating media.
Table 4 is an abridged table based upon Goodenough's values
for saturated ammonia. Tables 5 and 6, based upon Peabody's
Tables, show some of the properties of sulphur dioxide (S0 2 ) and
of carbon dioxide (C0 2 ).
PROBLEMS
1. Dry and saturated steam has a pressure of 100 lbs. per sq. in. absolute.
Calculate the temperature of the steam, the volume per pound, the heat
of the liquid, the latent heat, and the total heat above 32 F. per pound of
this steam.
2. Dry and saturated steam has a temperature of 300 F. What is
its pressure, heat of the liquid, latent heat, and total heat?
3. A closed tank contains 9 cu. ft. of dry and saturated steam at a
pressure of 150 lbs. per sq. in. absolute.
(a) What is its temperature?
(b) How many pounds of steam does the tank contain?
4. A boiler generates dry and saturated steam under a pressure of 200
lbs. per sq. in. absolute. The feed water enters the boiler at 6o° F.
(a) What is the temperature of the steam?
(b) How many British thermal units are required to generate
1 lb. of this steam if this feed water is admitted at 32 F.?
88 PROPERTIES OF VAPORS
(c) How many British thermal units are required to generate
i lb. of this steam from feed water at 60 ° F. into the
steam at the pressure stated at the beginning of this prob
lem?
5. One pound of dry and saturated steam is at a pressure of 250 lbs.
per sq. in. absolute.
(a) What is its internal energy of evaporation?
(b) What is its total internal energy above 32 F.?
(c) How much external work was done during its formation from
32° F.?
(d) How much external work was done during the evaporation?
(e) How much external work was done during the change in
temperature of the water from 32 degrees to the boiling
point corresponding to the pressure?
6. Dry and saturated steam is generated in a boiler and has a tempera
ture of 400 F. The feed water enters the boiler at 200 F.
(a) What pressure is carried in the boiler?
(b) What is the total heat supplied to generate 1 lb. of this steam?
(c) How much external work was done during its formation?
(d) How much heat was used in increasing the internal energy?
Check this by (hi — hi + II) noting that this assumes no
external work done in the heating of the liquid.
7. One pound of steam at a pressure of 100 lbs. per sq. in. absolute has
a quality of 90 per cent dry. What is its temperature?
8. How many British thermal units would be required to raise the
pound of steam in the above problem from 32 F. to the boiling point corre
sponding to the pressure stated?
9. What would be the volume of a pound of steam for the conditions
stated in problem 7?
10. How many heat units (latent heat) are required to evaporate the
steam in problem 7?
11. What is the amount of the total heat (above 32 F.) of the steam in
problem 7?
12. What would be the external work of evaporation of the steam in
problem 7?
13. How much external work (above 32 F.) is done in making steam as
in problem 7?
14. What is the internal energy of evaporation of the steam in problem 7?
15. What is the total internal energy of the steam in problem 7?
16. A tank contains 9 cu. ft. of steam at 100 lbs. per sq. in. absolute
pressure which has a quality of 0.95. How many pounds of steam does the
tank contain?
PROBLEMS 89
17. Two pounds of steam have a volume of 8 cu. ft. at a pressure of 100
lbs. per sq. in. absolute. What is the quality?
Calculate its total heat above 32 F.
18. One pound of steam having a quality of 0.95 has a temperature of
3 2 5 F. What is the pressure?
19. One pound of steam at a pressure of 225 lbs. per sq. in. absolute has
a temperature of 441. 9 F.
(a) Is it superheated or saturated?
(b) What is the total heat required to generate such steam from
water at 32 F.?
(c) What is its volume?
(d) How much external work was done (above 32 F.) in generat
ing it?
(e) How much internal energy above 32 F. does it contain?
20. One pound of steam at a pressure of 300 lbs. per sq. in. absolute has
a volume of 1.80 cu. ft.
(a) Is it saturated or superheated?
(b) What is its temperature?
(c) How much is its total heat above 32 F.?
(d) What is its total internal energy?
21. Steam in a steam pipe has pressure of 110.3 lbs. per sq. in. by the
gage. A thermometer in the steam registers 385 F. Atmospheric pres
sure is 14 lbs. per sq. in. absolute. Is the steam superheated, and if
superheated how many degrees?
22. Steam at a pressure of 200 lbs. per sq. in. absolute passes through
a throttling calorimeter.. After expansion into the calorimeter the tem
perature of this steam is 250 F. and the pressure 15 lbs. per sq. in. abso
lute. What is the quality?
23. Steam at a temperature of 325 F. passes through a throttling
calorimeter. In the calorimeter the steam has a pressure of 16 lbs. per
sq. in. absolute and a temperature of 236. 3 F. What is the quality?
24. Steam at 150 lbs. per sq. in. absolute pressure passes through a
throttling calorimeter. Assuming that the lowest condition in the calorim
eter for measuring the quality is io° F. superheat and the pressure in the
calorimeter is 15 lbs. per sq. in. absolute, what is the largest percentage of
wetness the calorimeter is capable of measuring under the above conditions?
25. Prepare a chart by means of which can be determined the largest
percentages of wetness a throttling calorimeter will measure at all pressures
from 50 to 300 lbs. per sq. in. absolute.
26. In a tenminute test of a separating calorimeter the quantity of dry
steam passing through the orifice was 9 pounds. The quantity of water
separated was 1 pound. What was the quality?
90 PROPERTIES OF VAPORS
27. A barrel contained 400 lbs. of water at a temperature of 50 F. Into
this water steam at a pressure of 125 lbs. per sq. in. absolute was admitted
until the temperature of the water and condensed steam in the barrel
reached a temperature of ioo° F. The weight of the water in the barrel
was then 418.5 lbs. What was the quality?
28. Calculate the factor of evaporation and the equivalent evaporation
per pound of coal for a boiler under the following conditions: Steam pres
sure, 190 lbs. per sq. in. gage; feed water temperature, 203 F.; steam appar
ently evaporated per pound of coal i\ pounds; steam 3 per cent wet.
29. Plot on one chart the pressuretemperature relations of ammonia,
SO2, and CO2.
30. Plot on one chart the pressurelatent heat relations of ammonia,
SO2, and CO2.
31. What conclusions do you reach from the graphic results in problems
29 and 30?
32. If cooling water for condensing ammonia cannot be secured at a tem
perature less than 90 F., what will be the pressure of the ammonia at the
inlet to the condenser of the refrigerating system?
CHAPTER VI
t
I
■3
w
SQ
/
dQ
T .i
Entropy 4>
 Analysis of Entropy
Diagram. •
ENTROPY
Pressurevolume diagrams are useful for determining the
work (in footpounds), done during any process or a cycle, but
they are of very limited use in analyz
ing the heat changes involved. It
has, therefore, been found desirable to
make use of a diagram which shows
directly by an area the number of
heat units (instead of footpounds) in
volved during any process. In order
that an area shall represent this value
the coordinates must be such that their
product will give heat units. If the
ordinates are in absolute temperature, Fig. 22.
the abscissas must be heat units per
degree of absolute temperature, that is, ^> for then T X^ = Q,
the amount of heat added during the process.
Suppose that the heat Q, which is required to produce a process,
such as an expansion or a compression, is divided up into a
number of small increments dQ (Fig. 22) and that each small
increment of heat is divided by the average absolute temperature
at which the heat change occurs. There will then be a series
of expressions ^ which, when integrated, will give the total
change in the abscissas. This quantity I ^ when multiplied by
the average absolute temperature between C and D will give the
total amount of heat involved during the process.
Mathematically expressed, the change in the abscissas is
dQ
T

v T
or <j> =
f
(121)
92 ENTROPY
The heat change involved is
dQ = Td<j> or Q = Ct d<t>. (122)
The quantity <j> in the equations is known as the increase in en
tropy of the substance, and may be denned as a quantity which,
when multiplied by the average absolute temperature occurring
during a process, will give the number of heat units (in B.t.u.)
added or abstracted as heat during the process. The " increase
in entropy" is employed rather than entropy itself, because
only the differences in entropy are important.
This definition of entropy states that in a temperatureentropy
diagram such as Fig. 22, where the ordinates are absolute tem
peratures, and the abscissas are entropies as calculated above
some standard temperature, the area under any line CD gives the
number of heat units added to the substance in passing from a
temperature T\ and entropy <£i = Oe to a temperature T 2 and
entropy <£ 2 = Of (or the number of heat units abstracted in pass
ing from T 2 to 7\) .
Entropy Changes During Constant Pressure Expansions of
Gases. The heat supplied or abstracted during a constant
pressure expansion or compression of a gas may be stated:
dQ = wCpdt, (123)
or (^ = wC p (r 2 — Ti). (124)
From equation (121) the change in entropy is
Assuming the specific heat C v constant, the change in entropy
becomes :
r T *dt
= wCp J  (126)
= WC P (loge T 2 — l0g e Ti)
= wC P \og e7 ^ (127)
I 1
ENTROPY CHANGES 93
Entropy Changes of Gases at Constant Volume. The heat
supplied or abstracted during a constant volume change of a gas
may be stated:
dQ = wCvdt, (128)
or Q = wC v (T 2 — Ti). (129)
Assuming, as before, the specific heat under constant volume
(C v ) constant and substituting in equation (121), the change in
entropy is
= wC v J  (130)
= WCv (l0g e T 2  lOge Ti) (131)
= wC v log e7 ^' (132)
1 1
Entropy Changes During Isothermal Processes of a Gas.
During an isothermal change the temperature remains constant.
The heat supplied or abstracted is consequently equal to the
heat equivalent of the work done, which in terms of weight and
temperature is
wRTloge^ ?  (133)
Vl
Since the temperature remains constant, the change in entropy
is
=
«* ri °*£; (134)
= WR \0g e ^. (135)
V 1
Entropy Changes During Reversible Adiabatic Processes of
Gases. The heat supplied to or abstracted from an adiabatic
expansion or compression is, by definition, zero. The entropy
change in such a process is
*=r?=° (i36)
Hence during an adiabatic process no change of entropy occurs.
"7
d 
s
T X A B
H 
T 2
D
c
(V
■»>
3
fH
o
OQ
.Q
e
/
94 ENTROPY
It is possible for adiabatic processes to occur which are not
reversible. These are processes in which, not only no heat is
added or abstracted, but a portion or all the work of the process
may reappear in the working medium as heat. In such a process
the entropy does not remain constant. Reversible adiabatic
processes are sometimes called isoentropic (equal entropy) to
distinguish them from the irreversible adiabatic processes.^
Entropy Changes During a Carnot Cycle. A pound of the
working substance is first expanded isothermally. On a Tcj>
(temperatureentropy) diagram (Fig.
23), this process would be represented
by line AB, where the temperature re
mains constant at T\, and where the
entropy increases from Oe to Of, be
cause of the addition of heat that is re
Entropy0 quired to keep the temperature constant.
Fig. 23. — Entropy Diagram The next process is adiabatic expan
of Carnot Cycle. • r t> 4. 't tt ^ • *i_i
3 sion from ii to 1 2  Heat is neither
added to nor abstracted from the substance during this expan
sion. Hence the entropy remains constant, as indicated by
BC.
The substance is now isothermally compressed along CD, the
temperature remaining constant at T 2 and the entropy decreasing
because of the abstraction of heat.
The last process of the cycle is the adiabatic compression from
D to A, no heat being added or abstracted, and the entropy,
therefore, remaining constant.
The heat supplied during the cycle is represented by the area,
ABfe and equals
($f  4>e) Tl (137)
The heat abstracted during the cycle is represented by the
area, CDef, and equals
(0/  e ) T 2 . (138)
Since the work of the cycle, in terms of heat, equals the heat
added minus the heat rejected,
TEMPERATUREENTROPY DIAGRAMS FOR STEAM 95
Work = Area ABfe — area CDef
= Area ABCD.
In terms of entropy,
Work = (</>/ — <f> e ) Ti — (</>/ — 4> e ) T 2
= (7\  T 2 ) (0/  e ). (139)
Efficiency of cycle is,
„ Area ABCD
E =
Area ABfe
(7\   T 2 ) (0/  4>c) = T,  T 2
T\ (0/ — 0e) 2\
(140)
From the foregoing discussion two important conclusions may
be drawn in regard to the use of the T<$> diagram :
1. If any heat process be represented by a curve on a T<f>
diagram, the heat involved during the process is equal to the
area under the curve, that is, between the curve and the axis of
absolute temperature.
2. If a cycle of heat processes be represented on a T0
diagram by a closed figure, the net work done is equal to the
enclosed area, that is, the enclosed area measures the amount
of heat that was converted into work.
T0 diagrams are useful for analyzing heat processes, and find
particularly useful application in steam engineering, as indicated
by the following:
1. Graphical analysis of heat transfers in a steam engine
cylinder.
2. Determination of quality of steam during adiabatic ex
pansion.
3. Calculations of steam engine cycle efficiencies.
4. Steam turbine calculations.
TemperatureEntropy Diagrams for Steam. A temperature
entropy diagram for steam is shown in Fig. 24. The various
shaded areas represent the heats added to water at 32 F. to
completely vaporize it at the pressure Pi. The area ABCD is
the heat added to the water to bring it to the temperature of
vaporization, or represents the heat of the liquid (h) for the pres
9 6
ENTROPY
sure P\. Further heating produces evaporation at the constant
temperature T\ corresponding to the pressure Pi, and is repre
sented by the area under the line CE. When vaporization is
complete, the latent heat, or the heat of vaporization (Z), is the
area DCEF. If, after all the water is vaporized, more heat is
added, the steam becomes superheated, and the additional heat
required would be represented by an area to the right of E.
Dry Steam or
Saturation Line
1.5
2.0
0.5 1.0
Entropy (0)
Fig. 24. — Temperatureentropy Diagram of Steam.
Calculation of Entropy for Steam. In order to lay off the
increase in entropy as abscissas in the heat diagram for steam,
it is necessary to determine various values. For convenience
3 2 F. has been adopted as the arbitrary starting point for cal
culating increase of entropy, as well as for the other thermal
properties of steam. Referring to Fig. 25 the entropy of water
is seen to be o at 32 F. In order to raise the temperature from
To to Ti, h heat units are required, and by the equation (121),
the entropy of one pound of the liquid 6 will be
C Tl dh ( ,
= / 57> 04i)
To
T
or, assuming the specific heat of water to be unity we can write
T
= l0ge T X  l0g e To = l0g e —
1
(142)
CALCULATION OF ENTROPY FOR STEAM
97
In order to evaporate the water into steam at the boiling point
Tij the latent heat, L, must be added at constant temperature
7\. The increase of entropy during the "steaming" process is
represented by — and is equal to, in this case,
Ora.
Eahr.
328+788
(143)
k ft
Entropy
logr e T 2 — H
log e Tx
Fig. 25. — Diagram for Calculation of Entropy of Steam.
Further heating would produce superheated steam and the
change in entropy would be
Entropy of superheat = /
= C p log
rsu p Cpdt
Ts&t. * '
T
J sup.
sat.
(144)
(145)
where C P = specific heat of the superheated steam at constant
pressure.
T S uv. = absolute temperature of the superheated steam.
Ts*t. = absolute temperature of the saturated steam.
98 * ENTROPY
Total Entropy of steam, which is the sum of entropies, corre
sponding to the various increments making up the total heat of
the steam, depends upon the quality of the steam.
For dry saturated steam the total entropy above 32 F. and
temperature 7\ is equal to the sum of the entropy of the liquid
and the entropy of evaporation for dry saturated steam. This
may be written
0sat. = #1 + — (l46)
J 1
= log e ^ + ^ (147)
i J 1
Similarly, for wet steam whose quality is x, the total entropy is
0wet = 01 + = (l48)
J 1
, 7\ xLi
J i 1
For superheated steam whose final temperature is T sup . and
the temperature corresponding to saturation is 7\, the total
entropy is, by equation (144),
*_* + £ +■/•*■*£* • ■ ( I49 )
1 1 «^r sa t. J
= loge ^ + £ + C p \0ge ^ ' (i 50)
1 2 J I JL sat.
Values of these entropies will be found in steam tables.
Referring to Fig. 25, the line Tic represents the increase of
entropy due to the latent heat added during the steaming process.
If this steaming process had stopped at some point such that
the steam was wet, having a quality x, this condition of the steam
could be denoted by the point s, where
This relation is obvious, for a distance along Tic represents
the entropy of steam, which is proportional to the latent heat
added, which in turn is proportional to the amount of dry steam
THE MOLLIER CHART 99
rrt
formed from one pound of water. In like manner, — ^— is the
T 2 d
quality of steam at the point m that has been formed along T 2 d
at the temperature T 2 .
It is thus apparent that any point on a Tcj> diagram will give
full information in regard to the steam. The proportional dis
tances on a line drawn through the given point between the
water and the dry steam lines and parallel to the Xaxis give
the quality of the steam as shown above. The ordinate of the
point gives the temperature and corresponding pressure, while
its position relative to other lines, such as constant volume and
constant total heat lines which can be drawn on the same dia
gram, will give further important data.
The Mollier Chart. While the temperatureentropy diagram
is extremely useful in analyzing various heat processes, its use.
in representing the various conditions of steam is not so conven
ient as that of the Mollier chart. This chart is illustrated in
Plate 1 of the Appendix. The coordinates consist of the total
heat of the steam above 32 F. and entropy. The saturation
curve marks the boundary between the superheated and satu
rated regions. In the wet region lines of constant volume are
drawn and in the superheated regions the lines of equal super
heats appear. In both the superheated and saturated regions,
lines of constant pressure are drawn and the absolute pressures
of the various curves are labeled.
By means of the Mollier chart (Appendix) the curve for
adiabatic expansion can be very readily drawn on a pressure
volume diagram when the initial quality of steam at cutoff
is known. Assume that one pound of wet steam at a pressure
of 150 pounds per square inch absolute and quality 0.965 dry
is admitted to the engine cylinder per stroke, and that there
is previously in the clearance space 0.2 cubic foot of steam
(see Fig. 26), at exactly the same condition. The volume of a
pound of dry saturated steam at this initial pressure of 1 50
pounds is, from the steamtables, 3.012 cubic feet. At 0.965
quality it will be 1 (0.965 X 3.012) = 2.905 cubic feet. On the
IOO
ENTROPY
scale of abscissas this amount added to the 0.20 cubic foot in
the clearance gives 3.105 cubic feet, the volume to be plotted at
cutoff. Other points in the adiabatic expansion curve can be
readily plotted after determining the quality by the method given
on page no.
^7 Clearance Steam
Admission
t Cutoff
0.20 3.15
Volume of Steam in Cylinder (Cu. ft.)
Fig. 26. — Illustrative Indicator Diagram of Engine using Steam with Expansion.
Temperatureentropy Diagram for the Steam Power Plant.
Fig. 27 illustrates the heat process going on when feed water
is received in the boilers of a power plant at ioo° F., is heated
and converted into steam at a temperature of 400 F., and then
loses heat in doing work. When the feed water first enters
the boiler its temperature must be raised from ioo° to 400 F.
before any " steaming" begins. The heat added to the liquid
is the area MNCD. This area represents the difference between
the heats of the liquid (374 — 68) or about 306 B.t.u. The
horizontal or entropy scale shows that the difference in entropy
between water at ioo° and 400 F. is about 0.436.*
The curve NC is constructed by plotting from the steam
tables the values of the entropy of the liquid for a number of
different temperatures between ioo° and 400 F.
If water at 400 F. is converted into steam at that temperature,
* As actually determined from Marks and Davis' Steam Tables (pages 9 and
15), the difference in entropy is 0.5663 — 0.1295 or 0.4368. Practically it is im
possible to construct the scales in the figure very accurately.
TEMPERATUREENTROPY DIAGRAM
IOI
the curve representing the change is necessarily a constant tem
perature line and therefore a horizontal, CE. Provided the
evaporation has been complete, the heat added in the " steaming "
process is the latent heat or the heat of evaporation of steam
(L) at 400 F., which is approximately 827 B.t.u.
400 •
•5100
§ 32!
u
1
C3
460
M
Supei heated
Steam
0.130
.566 1.528
Entropy — <P
Fig. 27. — Temperatureentropy Diagram for the Steam Power Plant.
The change in entropy during evaporation is, then, the heat
units added (827) divided by the absolute temperature at which
the change occurs (400 f 460 = 86o° F. absolute) or
L = 827 =
T 860
0.962.
The total entropy of steam completely evaporated at 400 F.
is, therefore, 0.566 + 0.962, or 1.528.* To represent this final
condition of the steam, the point E is plotted when entropy
measured on the horizontal scale is 1.528 as shown in the figure.
The point E is shown located on another curve RS, which is
determined by plotting a series of points calculated the same as
E, but for different pressures. The area MNCEF represents the
total heat added to a pound of feed water at 100 F., to produce
* Entropy, like the total heat (H) and the heat of the liquid (h), is measured
above the condition of freezing water (32 F.).
102 ENTROPY
steam at 400 F. Area OBCEF represents the total heat of
steam (H in the steam tables) above 32 F. required to form one
pound of steam at 400 F.
If the steam generated in the boiler is now allowed to expand
adiabatically in an engine cylinder or in a turbine nozzle to a
temperature of ioo° F., this expansion will be represented by the
line EG. GN represents the condensation of the exhaust steam.
The area NCEG represents the energy in the steam available
for work in the steam motor.
PROBLEMS
1. One pound of water is raised in temperature from 6o° to 90 F.
What is the increase in entropy?
2. Dry and saturated steam has a pressure of 100 lbs. per sq. in. absolute.
(a) Calculate the entropy of the liquid.
(b) Calculate the entropy of evaporation.
(c) Calculate the total entropy of the steam.
(d) Check values in (a), (b), and (c) with the steam tables.
3. Steam at 150 lbs. per sq. in. absolute has a quality of 0.90.
(a) What is the entropy of the liquid?
(b) What is the entropy of evaporation?
(c) What is the total entropy of the steam?
4. Steam having a temperature of 300 F. has an entropy of evapora
tion of 1. 1 900. What is its quality?
5. Steam having a pressure of 200 lbs. per sq. in. absolute has a total
entropy of 1.5400.
(a) What is the total entropy of dry and saturated steam under the
given pressure?
(b) Is the steam wet or dry?
(c) What is its quality?
6. Steam having a pressure of 125 lbs. per sq. in. absolute is super
heated ioo° F.
(a) What is the total entropy of dry and saturated steam under the
given pressure?
(b) What is the entropy of the superheat?
(c) What is its total entropy?
7. Steam having a pressure of 150 lbs. per sq. in. absolute has a total
entropy of 1.6043.
(a) What is the total entropy of dry and saturated steam under the
given pressure?
PROBLEMS 103
(b) Is the above steam saturated or superheated? How can you tell?
(c) How much superheat has the steam?
8. A boiler generates steam of 0.90 quality at a temperature of 350 F.
with the feed water admitted at oo° F. What is the increase in entropy?
9. Plot a temperatureentropy diagram for one pound of water vapor
for the pressures of 15, 50, 100, 150, 200, 250. The diagram should show the
liquid line, the 90 per cent quality line, the dry saturated line, and the 50
and ioo° superheat curves for each of the given pressures.
10. What is the entropy of steam 92 per cent dry at a pressure of 15 lbs.
per sq. in. absolute?
n. With a quality of 0.90, what is the entropy of evaporation of steam
at a pressure of 25 lbs. per sq. in. absolute?
12. What is the total entropy of steam 94 per cent dry at a pressure of
100 lbs. per sq. in. absolute?
CHAPTER VII
EXPANSION AND COMPRESSION OF VAPORS
Vapors, like gases, can be expanded or compressed, but the
laws governing their thermodynamic relations are more complex.
In the heat changes of vapors the heat that is required to change
the state of the substance must be accounted for, in addition to
that necessary to do the external work and to produce the change
in temperature of the substance.
Equation 12, showing the effect of the heat added or abstracted
during an expansion or a compression, applies equally to vapors
and gases.
The external work (equation 1 2) done during the expansion or
compression of vapor is calculated, as in the case of gases, from
the area under the expansion or compression curve. Work,
being a product of pressure and volume, is independent of the
working medium.
The internal energy changes (equation 12) involved during an
expansion or compression of vapors are best measured by dif
ferences. The internal energy at the beginning and at the end
of an expansion or compression may be determined by reference
to the vapor tables. The values, as obtained from such tables
(Tables 3, 4, 5, 6 in Appendix), are calculated from a standard
datum temperature of 32 F., and their difference gives the change
in internal energy involved in the process in question.
Thus, for any vapor the following general equation may be
written:
Heat added = Internal energy at the end of the expansion or
compression minus the internal energy at the beginning of the
expansion or compression plus the heat equivalent of the external
work done, or
Q = (h h) + W. (151)
104
EXPANSION OF WET STEAM AT CONSTANT VOLUME 105
The following problems show the application of equation (151)
to various types of expansions.
1. Expansion of Wet Steam at Constant Volume. Assume
that one pound of steam at a pressure of 15 pounds per square
inch absolute, and 50 per cent dry, receives heat under constant
volume raising the pressure to 30 pounds per square inch absolute.
Find: (a) the volume of the steam after the addition of the
heat, (b) the quality of the steam, (c) the work done, (d) the
heat added.
Solution: (a) Since the volume remains constant the final
volume of the steam is
xFsat. = 0.50 X 26.27 = 13.13 cu. ft.,
in which x = the quality of the steam, V = volume of saturated
steam at 15 pounds per square inch absolute pressure (26.27)
as obtained from steam tables, in cubic feet per pound.
(b) From steam tables the volume of one pound of dry and
saturated steam at the final condition of 30 pounds per square
inch absolute is 13.74. Since the actual volume is less than that
of the saturated steam, the steam in its final condition is wet and
the quality is
from which
x = ■ = 0.955 or 955 P er cent dry.
(c) Since the volume is constant the work done is
W = o.
(d) The heat added is
Q = h  h + W
= h  h + o
= h  h.
The internal energy at the final condition is, from equation
(no),
h = h+X2 \L 2 * — L \ = h 2 + X2P2
778
106 EXPANSION AND COMPRESSION OF VAPORS
= (218.8 + O.955 X 869.0)
= 218.8 + 829.9 = 1048.7 B.t.u.,
where fa and p 2 are the heat of the liquid and internal latent
heat respectively at 30 pounds per square inch absolute pressure.
The internal energy at the initial condition is
Ii =\ h + Xipi
= (181.0 + 0.50 X 896.8)
= (181.0 + 448.4) = 629.4 B.t.u.
Then,
Q = 1048.7 — 629.4
= 419.3 B.t.u.
2. Expansion of Superheated Steam at Constant Volume.
One pound of steam at 130 pounds per square inch absolute pres
sure and 50 F. superheat is heated under constant volume to
a pressure of 180 pounds per square inch absolute.
Find (a) the final quality of the steam;
(b) the heat supplied.
Solution: (a) the initial volume of the steam is found by
reference to the superheated steam tables to be 3.74 cubic feet.
Since this equals the final volume of the steam, the final condition
of the steam is 300 F. superheat, which from the steam tables is
the condition when steam at 180 pounds per square inch absolute
pressure has a volume of 3.74 cubic feet.
(b) The heat supplied equals
Q = h  h + W
= J1/1.
No values corresponding to the internal energy are available
in the superheated tables and consequently these must be cal
culated.
*  ^) to)
where
(■
#sup. = the total heat of superheated steam as found in the
superheat tables.
EXPANSION AT CONSTANT PRESSURE 107
P 2 = pressure in pounds per square foot.
V 2 = volume of the superheated steam, cubic feet per pound.
Thus, for the conditions of the problem,
T ( 180 X 144 X 374\
h = (13539 jfe^)
= 1 3539 ~ 124.6 = 1229.3 B.t.u.
t ( I 3° X x 44 X 374\
/ 1= ^ 2I9 . 7 _3 ^—^±)
= 1219.7 — 89.5 = 1130.2 B.t.u.,
from which
Q = 1229.3 — 1130.2 = 99.1 B.t.u.
3. Expansion at Constant Pressure. One pound of steam
at a pressure of 150 pounds per square inch absolute and a
volume of 1.506 cubic feet expands under constant pressure until
it becomes dry and saturated.
1. What is the quality at the initial condition?
The volume of dry and saturated steam at the given pres
sure is 3.012 cubic feet per pound.
The quality then is — — = 0.50.
3.012
2. What is the volume of the steam at the final condition?
The volume of a pound is 3.012 cubic feet since the steam
is dry and saturated.
3. What is the work done during the expansion?
Work = Pi (V 2  Vi) = 150 X 144 (3.012  1.506) = 32,530
footpounds.
4. How much heat is required? Several methods of reason
ing may be used in calculating the values of the heat required
and the work done during constant pressure expansion or
compression.
Q = h  h + W
h = fa + X2P2
= 330.2 + I X 780.4 = 1 1 10.6
108 EXPANSION AND COMPRESSION OF VAPORS
1 1 = hi + Xipi
= (330 2 + 0.50 X 780.4)
= 720.4 B.t.u.
The heat equivalent of the external work during the expansion
is
^ = ^ i2 C53)
150 X 144 (1 X 3.012  0.50 X 3.012)
= = A.1.0.
778
Substituting in equation (151),
Q = 1110.6 — 720.4 + 41.8
= 432.0 B.t.u.
The amount of heat required in the case of constant pressure
expansion may be calculated directly:
Q = (x 2  Xi) L, (154)
where x 2 and Xi equal the qualities at the initial and final con
ditions of the expansion; L equals the latent heat of dry and
saturated steam at the given pressure. Thus, for the above
problem,
Q = (1.00 — 0.50) 863.2 = 431.6 B.t.u.
Similarly, the heat supplied during constant pressure expansion
may be expressed:
Q = H 2  H h
where H 2 and Hi are the total heat of the steam above 32 for
the initial and final conditions respectively. Thus, for the prob
lem in question,
H 2 = 330.2 + ioo X 863.2
Hi = 330.2 + 0.50 X 863.2
Q = (330.2 + 1.00 X 863.2)  (330.2 + 0.50 X 863.2)
= 431.6 B.t.u.
Isothermal Lines for Steam. When the expansion of steam
occurs at constant pressure as, for example, in the conversion
ADIABATIC LINES FOR STEAM 109
of water into steam in a boiler when the engines are working, we
have isothermal expansion. Isothermal lines for wet steam,
which consists of a mixture of water and its vapor, are, there
fore, straight lines of uniform pressure. On a pressurevolume
diagram an isothermal line is consequently represented by a
horizontal line parallel to the axis of abscissas. As steam
becomes superheated the pressure decreases as the volume
increases; for highly superheated vapors the isothermal curve
approaches a rectangular hyperbola. On a Tcf) diagram, the
isothermal line is represented by a line of constant temperature,
i.e., by a line parallel to the <£axis.
Adiabatic Lines for Steam. Adiabatic lines will have differ
ent curvature depending upon the substances used, It will be
remembered that the values of 7 are different for the various
gases and therefore the adiabatic line for each of these gases
would have a different curvature. In the same way the curva
ture of adiabatic lines of steam will vary with the quality of the
steam. Steam which is initially dry, if allowed to expand adia
batically, will become wet, the percentage of moisture which it
will contain depending on the extent to which the expansion is
carried. Also, on any T<j> diagram, an adiabatic (isentropic) line
is represented by a line parallel to the Taxis, i.e., by a line of
constant entropy. If steam is initially wet and is expanded adia
batically, it becomes wetter as a rule.*
In general, in any expansion, in order to keep steam at the
same relative dryness as it was initially, while it is doing work,
some heat must be supplied. If the expansion is adiabatic so
that no heat is taken in, a part of the steam will be condensed
and will form very small particles of water suspended in the
steam, or it will be condensed as a sort of dew upon the surface
of the enclosing vessel.
The relation between pressure and temperature as indicated
by the steamtables continues throughout an expansion, pro
vided the steam is initially dry and saturated or wet.
* When the percentage of water in steam is very great and the steam is ex
panded adiabatically there is a tendency for the steam to become drier. This
is very evident from an inspection of diagrams like Fig. 25.
HO EXPANSION AND COMPRESSION OF VAPORS
Adiabatic Curve for Steam. Whether steam is initially dry
and saturated or wet, the adiabatic curve may be represented
by the formula: PV n = constant. The value of the index n
depends on the initial dryness of the steam. Zeuner has deter
mined the following relation:
n = 1.035 + — (i55)
10
Solving this when x = unity (dry and saturated steam) the
value of n is 1.135, and when x is 0.75, n has the value 1.11.*
While the work during an adiabatic expansion may be cal
culated from the value of n as determined from Zeuner's relation
(equation 155), the following method is more commonly used.
Since by definition the heat added or abstracted during an
adiabatic process is zero, the heat equation becomes,
Q = I 2 h + W = 0,
or W = h  1 2 (156)
The work done by an adiabatic process can then be determined
by the difference between the internal energies at the beginning
and end of the process.
In order to calculate the internal energies the qualities must
be known and these may be found from the entropies.
Quality of Steam During Adiabatic Expansion. Since in an
adiabatic expansion no heat transfer takes place, the entropy
remains constant, and, therefore, on a T<j> diagram, this condition
is represented by a straight vertical line as cgf (Fig. 25) or smk.
Thus for wet steam:
cf> = 6 + x, (157)
where x = the quality or dryness fraction of the steam, <j> =
* Rankine gave the value of n = V, which obviously from the results given
is much too low if the steam is at all near the dry and saturated condition. His
value would be about right for the condition when x = 0.75. In an actual steam
engine, the expansion of steam has, however, never a close approximation to the
adiabatic condition, because there is always some heat being transferred to and
from the steam and the metal of the cylinder and piston.
QUALITY OF STEAM DURING ADIABATIC EXPANSION III
the total entropy of saturated steam, and = the entropy of
the water.
Since in adiabatic expansion the entropy remains constant,
the following equation can be written
0i = 02 (total entropies)
or
Bi + xM = e 2 + xM (i 5 8)
il 1 2
Knowing the initial conditions of steam, the quality of the
steam at any time during adiabatic expansion can be readily
determined* Thus, suppose the initial pressure of dry satu
rated steam to be 100 pounds per square inch absolute, and the
final pressure after adiabatic expansion 1 7 pounds per square inch.
From the steam tables we find that the total entropy 0i, for
dry steam at 100 pounds pressure, is 1.6020; that is
0i = 1.6020.
The entropies at 17 pounds pressure are also obtained from the
tables,
1.6020 = 0.3229 f # 2 14215,
whence
#2 = 0.899.
For a rapid and convenient means of checking the above
result, the " Total Heatentropy " diagram in the Appendix can
be used. From the intersection of the 100pound pressure line
and that of unit quality (" saturation line") is dropped a vertical
line (line of constant entropy = 1.602) to the 17pound pressure
line. This latter intersection is found to He on the 0.90 quality
line.
Example. One pound of steam having a quality of 0.95 at a
pressure of 100 pounds per square inch absolute expands adia
batically to 15 pounds per square inch absolute.
1. What is the quality at the final condition?
The total entropy at the initial condition equals
+ x = 0.4743 + 0.95 x 1. 1277 = 1.5456.
112 EXPANSION AND COMPRESSION OF VAPORS
The total entropy at the end of the expansion equals
+ x  = 0.3133 + 1.4416 x.
Since entropy is constant in adiabatic expansion
0.3133 + 1.4416 s = i.545 6 >
from which the final condition is: x = 0.854.
2. How much work is done during the expansion?
Since there is no heat added the work done equals the loss in
internal energy.
The internal energy at the end of the expansion equals
h + X2P2 = 181. o + 0.854 X 896.8 = 946.9 B.t.u.
The internal energy at the initial condition equals
h + Xipi = 298.3 + 0.95 X 806.6 = 1064.6 B.t.u.
The work equals
1064.6 — 946.9 = 1 1 7.7 B.t.u. or 91,576 footpounds.
Poly tropic Expansion {n — 1).
Example. One pound of steam has a pressure of 100 pounds
per square inch absolute and a quality of 0.95. It expands
along an n = 1 curve to 20 pounds per square inch absolute.
What is the quality at the end of the expansion?
Solution. The volume of the steam at the initial condition is
X X F S at.
x X F S at. = 0.95 X 4.429 = 4.207 cubic feet.
Obviously, PxV? = P 2 V 2 n
and since n — 1,
100 X 4.207 = 20 X V2,
V 2 = 21.035 cubic feet.
The volume of dry saturated steam at the end of the expan
sion or at 20 pounds per square inch is 20.08. Therefore the
steam is superheated at end of expansion. How is this known?
GRAPHICAL DETERMINATION OF QUALITY OF STEAM 113
Graphical Determination of Quality of Steam by Throttling
Calorimeter and Total Heatentropy Diagram. It will be re
membered that the throttling calorimeter (pages 77 to 81) de
pends for its action upon the fact that the total heat of steam
which expands without doing work remains the same, the heat in
excess of that required to keep the steam dry and saturated going
to superheat the steam. Suppose that steam enters the calorim
d.50
1.60
1.70
1.80
Entropy
Fig. 28. — Mollier Diagram for Determining Quality of Steam.
eter at a pressure of 150 pounds per square inch absolute, and is
throttled down to 1 7 pounds per square inch, the actual tempera
ture being 24o ? F. Since the saturation temperature for steam
at 17 pounds pressure is 219.4, the steam in the calorimeter is
superheated 240 — 219.4 or 20.6 degrees. In order to find the
quality of the live steam refer to the " Mollier Diagram " (Appen
dix or Fig. 28) and find the intersection of the 20.6 degrees super
heat line with the 1 7pound pressure line. From this point follow
114 EXPANSION AND COMPRESSION OF VAPORS
a horizontal line (line of constant total heat) to the left until it
intersects the 150pound pressure line. This point of intersec
tion is found to lie on the 0.96 quality line.
Formula 117 (page 80) gives the following result in close
agreement with the diagram:
_ 1153.1 + 047 (240 ~ 2194)  330.2
Xl ~ 863.2
= 0.965.
PROBLEMS
1. Dry saturated steam at 100 lbs. per sq. in. absolute pressure contained
in a closed tank is cooled until its pressure drops to 15 lbs. per sq. in. abso
lute. What is the final quality and the heat removed from each pound of
steam?
2. One pound of steam at 15 lbs. per sq. in. absolute has a volume of 1 2.36
cu. ft. It is heated under constant volume until the pressure becomes
50 lbs. per sq. in. absolute, (a) What is the quality before and after the
heating? (b) How much heat was supplied?
3. A closed tank containing dry and saturated steam at 15 lbs. per sq. in.
absolute pressure is submerged in a body of water at a temperature of 59 F.
What will be the ultimate pressure and quality of steam within the tank?
4. Prove that for a constant pressure expansion the heat supplied equals
the difference between the total heats of the vapor at the beginning and at
the end of the expansion.
5. One pound of steam at 100 lbs. per sq. in. absolute pressure and 50 per
cent dry expands at constant pressure. What work is done and what heat
is required to double the volume? "What is the temperature at the beginning
and at the end of the expansion?
6. One pound of steam at a pressure of 100 lbs. per sq. in. absolute
and 50 per cent dry is expanded isothermally until it is dry and saturated.
Find the heat supplied and the work done.
7. One pound of steam at a pressure of 150 lbs. per sq. in. absolute has a
quality of 0.90. What work is done and what heat is required to double
its volume at constant pressure?
8. If steam at 200 lbs. per sq. in. absolute, 95 per cent dry, is caused to
expand adiabatically to 228 F., what are the properties of this steam at
the lower point? (That is, final total entropy, entropy of evaporation,
quality and volume.)
9. What will be the final total heat of dry saturated steam that is ex
panded adiabatically from 150. lbs. per sq. in. absolute down to 10 lbs.
per sq. in. absolute?
PROBLEMS 115
10. Steam having a quality of 0.20 dry is compressed along an adiabatic
curve from a pressure of 20 lbs. per sq. in. absolute to a pressure correspond
ing to a temperature of 293 F. What is the final quality?
11. Determine the final quality of the steam and find the quantity of
work performed by 2 lbs. of steam in expanding adiabatically from 250 lbs.
per sq. in. absolute pressure to 100 lbs. per sq. in. absolute, the steam being
initially dry and saturated.
12. One pound of steam at 150 lbs. per sq. in. absolute, and 200 F.
superheat, expands adiabatically. What is the pressure when the steam
becomes dry and saturated? What is the work done during the expan
sion?
13. One pound of dry and saturated steam at 15 lbs per. sq. in. absolute
pressure is compressed adiabatically to 100 lbs. per sq. in. absolute pressure.
What is the quality at the end of the compression and the negative work
done?
14. One pound of steam at 100 lbs. per sq. in. absolute pressure has a
quality of 0.80. It expands along an n = 1 curve to a pressure of 15 lbs.
per sq. in. absolute, (a) What is the volume at the beginning and at the
end of the expansion, (b) the quality at the end of the expansion, (c) the
work done during the expansion, (d) the heat supplied to produce the expan
sion?
15. One pound of steam at 150 lbs. per sq. in. absolute expands along an
n = 1 curve to 15 lbs. per sq. in. absolute. The quality of the steam at
15 pounds is to be dry and saturated, (a) What must be the quality at the
initial conditions? (b) What work will be done by the expansion? (c) What
heat must be supplied?
16. Steam at a pressure of 100 lbs. per sq. in. absolute having a quality
of 0.50 expands adiabatically to 15 lbs. per sq. in. absolute. What is the
quality at the end of the expansion?
17. One pound of steam at a pressure of 100 lbs. per sq. in. absolute has
a volume of 4 cu. ft. and expands adiabatically to 15 lbs. per sq. in. absolute.
(a) What is the quality at the initial and final conditions?
(b) What is the work done during the expansion?
18. One pound of steam having a pressure of 125 lbs. per sq. in abso
lute and volume of 4.17 cu. ft. expands adiabatically to 25 lbs. per sq. in.
absolute.
(a) What is the quality at the initial and final conditions?
(b) What is the work of expansion?
19. Given the steam as stated in problem 18 but with expansion complete
at 100 lbs. per sq. in. absolute. What is the quality at this pressure?
20. What would be the pressure if the steam in problem 18 were expanded
adiabatically until it became dry and saturated?
n6 EXPANSION AND COMPRESSION OF VAPORS
21. One pound of steam at a temperature of 360 F. has a quality of
0.50, and expands under constant pressure to a volume of 3.4 cu. ft.
(a) What is the quality at the final condition?
(b) What is the work of the expansion?
(c) What heat is required?
22. Two pounds of steam at a pressure of 100 lbs. per sq. in. absolute
have a volume of 4 cu. ft., and expand under constant temperature to a
volume of 8 cu. ft.
(a) What is the quality at the initial and final conditions?
(b) What is the work of the expansion?
(c) How much heat is required?
CHAPTER VIII
CYCLES OF HEAT ENGINES USING VAPORS
Carnot Cycle. The Carnot cycle using a vapor employs the
same apparatus as was explained on page 40. The cycle is made
up of two isothermals and two adiabatics, but differs from the
cycle using gas as the working medium in that the isothermal
curves are lines of constant pressure.
Volume
Fig. 29. — Camot Cycle using Vapors.
As an illustration, assume that the vapor in the cylinder is in
the liquid state and at a temperature T\ when the heat is applied.
The heating process continues until the vapor becomes dry and
saturated. In order to maintain the temperature constant the
heat must be supplied at such a rate as to maintain the pressure
constant. The volume increases during this change from that
of the specific volume of the liquid to the specific volume of the
dry saturated vapor at the temperature TV This change is
represented by the lines ah on the PV and temperatureentropy
diagrams, Figs. 29 and 30.
117
n8
CYCLES OF HEAT ENGINES USING VAPORS
The cylinder is then removed from the source of heat and the
adiabatic expansion is produced. This is represented in Figs.
29 and 30 by the curves be.
The isothermal compression curve cd is produced when the
cylinder is placed in communi
cation with the refrigerator or
condenser and heat is absorbed
from the vapor. Since the cyl
inder contains a saturated va
por, the process results in a
constant pressure curve at tem
perature, r 2 .
The adiabatic compression
curve da follows when the cyl
inder is removed from the re
frigerator. The vapor in its
final condition is reduced to a
liquid at the temperature 7\.
The heat added to the cycle
is that required to produce the
line ab, Figs. 29 and 30. For the assumed case this is:
<2i = H b  h a = L h (159)
where H b = total heat of dry saturated steam at the pressure P b ,
h a = heat of liquid above 32 F. for condition at a.
The heat rejected from the cycle is that absorbed by the
refrigerator and for each pound of vapor it is equal to
Q2 = (h c + XcL c ) — {h d + XdLa). (160)
From the temperatureentropy diagram, Fig. 30, the heat added
and heat rejected are
Qi = T x (fa  <£«), (161)
ft = T 2 (fa  0d). (162)
Since (</> c — </><*) = (<£& — <t>a),
Q 2 = T 2 (</>»  4> a ). (163)
Entropy^
Fig. 30. — Temperatureentropy Diagram
of Carnot Cycle using Vapors.
THE RANKINE CYCLE 119
The work of the cycle may be determined from the algebraic
sums of the work under the individual curves, but can be more
simply calculated from the temperatureentropy diagram, Fig.
30. The area abed is proportional to the work and since it is a
rectangle,
work of cycle = (T 2 — 7\) (<fr> — <£ a ).
From equations (161) and (163), the efficiency of the cycle is
77 _ (T2 — Ti) ((f> b — a )
T\ (</>& — <f>a)
T 2 T x
(164)
Equation (164) shows that the efficiency of Carnot's cycle is
not affected by the character of the working substance and is
dependent only upon its initial and final temperatures.
The Rankine Cycle.* The Carnot cycle gives the maximum
efficiency obtainable for a heat engine operating between given
limits of temperature. In order that a steam engine may work on
a Carnot cycle, the steam must be evaporated in the cylinder
instead of in a separate boiler, and condensed in the cylinder
instead of being rejected to the air or to a separate condenser.
Such conditions are obviously impracticable, and it has, therefore,
been found necessary to adopt some other cycle which conforms
more with practical conditions. The most efficient practical
steam cycle, and the one which has been adopted as the standard
with which the efficiency of all steam engines may be compared
is the Rankine Cycle. The pressure volume diagram of this cycle
is shown in Fig. 31. Steam is admitted at constant pressure and
temperature along ab to a cylinder without clearance. At b cut
off occurs, and the steam expands adiabatically from b to c, some
of i condensing during the process. The steam is then dis
charged at constant pressure and temperature along the back
pressure line cd. Line da represents the rise in temperature and
* Also known as the Clausius Cycle, having been published simultaneously
and independently by Clausius.
120
CYCLES OF HEAT ENGINES USING VAPORS
pressure at constant volume when the condensed steam is con
verted into a vapor.
The four stages of the Rankine cycle may be stated as follows:
(i) Feed water raised from temperature of exhaust to tem
perature of admission steam. (Line da.)
(2) Evaporation at constant admission temperature. (Line
ab.)
(3) Adiabatic expansion down to back pressure. (Line be.)
(4) Rejection of steam at the constant temperature corre
sponding to the back pressure. (Line cd.)
Volume
Fig. 31. — Indicator Diagram of Ideal Rankine Cycle.
The net work done in the cycle, assuming one pound of wet
steam, is calculated as follows:
1. External work of evaporation (in B.t.u.),
(W») = tIs (PiFi*i). (165)
2. Loss in internal energy,
(W bc ) = fa + Km — (fa + X2P2) B.t.u. (166)
3. External work of evaporation at temperature of exhaust
(B.t.u.),
W dc =  Thf (PcVc) =  T^8 (P2V 2 X 2 ). (167)
4. W da  O. (168)
THE RANKINE CYCLE 121
Adding,
Net work of cycle,
W = ffaPiViXi + hi + xipi  h x 2 p 2
— TT 8" P2 V 2 X 2 ,
but
•yyg P^V\ + pi — £1.
Therefore,
Tf = h + X1Z1  (fe + x 2 L 2 ) (B.t.u.). (169)
Thus the net work of the Rankine cycle is equal to the dif
ference between the total heat of the steam admitted and the
total heat of the steam exhausted. This statement applies
whether the steam is initially wet, dry or superheated.
The heat added during the cycle is
Qi = h a + XbLj, — h d . (170)
The heat rejected during the cycle is
Q 2 = XcLc. (171)
The efficiency of the Rankine cycle is
E __ Qi— Q2 = h a + x b L b — ha — XcLc t
Qi h a + x^ — h d
Since h a = h b and h c = h d ,
77 f^b ~~ X^L,^ He XcL/c / \
rt b j XfrL b lie
Fig. 32 is the T<j> diagram for a Rankine cycle using dry
saturated steam. The letters abed refer to the corresponding
points in the pressure volume diagram Fig. 31. The net work
of the cycle is the area B + C, which is the difference between
the total heats at admission and exhaust. The heat added per
cycle is represented by the areas A + B + C + D and the
B + C
Thermal efficiency = A + B + Q + D (173)
hi + X\Li — h 2 — x 2 L 2 ( \
= j — ; j 7 ^74;
hi + X1L1 — rh
* The final quality can be determined by equation (158).
122
CYCLES OF HEAT ENGINES USING VAPORS
fe in the denominator must always be subtracted from hi + X\L\
(the total heat above 3 2° F.), in order to give the total heat above
the temperature of feed water, which in engine tests is always
assumed for the purpose of comparison to be the same as the
exhaust temperature.
Entropy—
Fig. 32. — Temperatureentropy Diagram of Rankine Cycle.
Example. One pound of steam at a pressure of 160 pounds per
square inch absolute and quality of 0.95 performs a Rankine cycle
exhausting at 5 pounds per square inch absolute.
1. What is the quality of the exhaust?
The total entropy at the initial condition
= 0.5208 + 0.95 X 1.0431.
The total entropy at the exhaust
= 0.2348 + 1.608 x.
Then 0.5208 + 0.95 X 1.0431 = 0.2348 + 1.6084 #.
From which x = 0.794.
2. What is the net work of the cycle?
Work = Hi  H 2 = 335.6 + 0.95 X 858.8  130.1  0.794
x 1000.3 = 227.2 B.t.u.
3. What is the efficiency of the cycle?
THE RANKINE CYCLE
123
Efficiency =
227.2
335.6 + 0.95x858.8130.1
= 0.222 or 22.2 per cent.
In the actual steam engine cycle it is impractical to expand the
steam down to the backpressure line.
The Rankine cycle for the actual steam engine is similar to
that described, except that the adiabatic expansion terminates at
a pressure higher than that of the backpressure, that is, the
expansion is incomplete.
The pressurevolume and T4> diagrams of this modified
Rankine cycle using dry saturated steam are shown in Figs. 33
and 34. The cylinder is without clearance. Steam is admitted
Volume
Fig. 33. — Rankine Cycle, Incomplete Expansion.
at constant pressure and temperature along ab. Cutoff occurs
at point b and the steam expands adiabatically to the terminal
pressure c. Part of the steam is discharged at constant volume
cd. The remainder is exhausted during the backpressure stroke
de. Line ea represents the rise in temperature of the feed water
from T 2 to 2\.
The net work of the cycle can best be calculated by dividing
Fig* 33 into the two areas abed and e'ede. The area abed is that
of the theoretical or ideal Rankine cycle, while dede is that of the
rectangle. The net work of the cycle is equal to the sum of these
two areas.
124
CYCLES OF HEAT ENGINES USING VAPORS
The heat equivalent of the area abed is
hb + XbLb — h c — XcL c .
The area e'ede is
Tb (Pc  Pa) (Va  o).
The total heat equivalent of the work of the cycle is
(Pc  Pa) Va
h + XbLb — h c — XcLc +
The heat added to the cycle is
hb + XbLb — ha.
The efficiency of the cycle is
778
hbXbLb — h c —XcLc +
E =
{Pc  Pa) Va
778 *
hb + XbLb — ha
(i75)
(176)
(i77)
(178)
(i79)
In the T<j> diagram, Fig. 34, the letters abede refer to
corresponding points in the pressurevolume diagram (Fig.
33). The net work of the cycle is represented by the area B
+ C. The heat added to the
cycle is represented by the area
A + B + C + D.
The incomplete Rankine cy
cle (Figs. 33 and 34) is less
efficient than the ideal cycle
(Figs. 31 and 32), because of
failure to expand the steam
completely. This loss is repre
sented in Figs. 33 and 34 graphi
cally by the area cdf. This
cycle is sometimes used as a
standard in preference to the
ideal Rankine cycle when the
efficiencies of engines are com
pared.
Example. One pound of steam at a pressure of 160 pounds
per square inch absolute and quality of 0.95 goes through a
Entr.opy0
Fig. 34. — Rankine Cycle, Incomplete
Expansion.
THE PRACTICAL OR ACTUAL STEAM ENGINE CYCLE 125
Rankine cycle. The terminal pressure is 15 pounds per square
inch absolute and the exhaust pressure 5 pounds per square inch
absolute. What is (a) the work of the cycle, (b) the heat added
to the cycle, and (c) the efficiency of the cycle?
Solution. The total entropy of the steam at the initial con
dition is
0.5208 + 0.95 X 1.0431.
The total entropy of the steam at the terminal pressure is
0.3133 + x (1.4416).
Then 0.5208 + 0.95 X 1.0431 = 0.3133 + x (1.4416)
from which the quality at the terminal pressure is
x = 0.831.
The volume of the steam at the terminal pressure is
0.831 X 26.27 = 21.83.
The total heat of the steam at the initial condition is
335.6 + 0.95 X 858.8 = 1151.5.
The total heat of the steam at the terminal pressure is
181.0 + 0.831 X 969.7 = 986.8.
The work of the cycle is then (in terms of heat units),
Trr on o . ( T 5 ~ 5) X X 44 X 21.83
w = 1151.5  986.8 + ^2 — D/ r* * = 205.1.
778
The heat added to the cycle is
335.6 + 0.95 x 858.8 — 130.1 = 1021.4 B.t.u.
The efficiency of the cycle is
= 20.08 per cent.
1 02 1. 4
The Practical or Actual Steam Engine Cycle. In the steam
engine designed for practical operation it is impossible to expand
the steam down to the backpressure line; and, furthermore, it
126
CYCLES OF HEAT ENGINES USING VAPORS
is evident that some mechanical clearance must be provided.
The result is that in the indicator diagram from the actual steam
engine, we have to deal with a clearance volume, and both incom
plete expansion and incomplete compression as shown in Fig.
35. In order to calculate the theoretical efficiency of this prac
9
b
c\
h
a
%
[
83
\
<o
\
Si
\
Ph
\
i
3
\
Sl_
 '!»#
Xf
Volume
Fig. 35. — Indicator Diagram of Practical Engine Cycle.
tical cycle, it is necessary to assume that the expansion line cd
and the compression line fa are adiabatic. Knowing then the
cylinder feed of steam per stroke and the pressure and volume
relations as determined from the indicator diagram, one can
calculate the theoretical thermal efficiency by obtaining the net
area of the diagram (expressed in B.t.u.) and dividing by the
heat supplied per cycle. In order to obtain the net area of the
diagram, the latter may be divided up into several simple parts
as follows:
gcdi — area equivalent to a Rankine cycle,
idej — a rectangle,
hafj — area equivalent to a Rankine cycle (negative),
gbah — a rectangle (negative).
The actual efficiency of the steam engine is usually determined
by dividing the heat equivalent to a horse power by the heat in
the steam required to produce a horse power. Since one horse
AN ENGINE USING STEAM WITHOUT EXPANSION 1 27
power per hour is equal to ^M^L or 2545 B.t.u., the actual
efficiency of a steam engine is
E =
2545
WR(Hh) (l8o)
In equation (180) WR is the water rate or the steam consump
tion per horse power per hour, H is the total heat in the steam
at the initial pressure and quality as it enters the engine, h is the
heat in the feed water corresponding to the exhaust pressure.
Efficiency of an Engine Using Steam Without Expansion. In
the early history of the steam engine, nothing was known about
the " expansive " power of steam. Up to the time of Watt in
all steam engines the steam was admitted at full boiler pressure
at the beginning of every stroke and the steam at that pressure
carried the piston forward
to the end of the stroke
without any diminution of
pressure. Under these cir
cumstances the volume of
steam used at each stroke
at boiler pressure is equal to
the volume swept through
b}r'the piston.
An indicator diagram rep
resenting the use of steam
in an engine without expan
sion is shown in Fig. 36.
This diagram represents
steam being taken into the engine cylinder at 1 at the boiler
pressure. It forces the piston out to the point 2 when the
exhaust opens and the pressure drops rapidly from 2 to 3.
On the back stroke from 3 to 4 steam is forced out of the
cylinder into the exhaust pipe. At 4 the pressure rises rapidly
to that at 1 due to the rapid admission of fresh steam into
the cylinder. In this case the thermal efficiency (E) is repre
sented by
— s> Volume
Fig. 36. — Indicator Diagram of an Engine
using Steam without Expansion.
128 CYCLES OF HEAT ENGINES USING VAPORS
E = work done = (P x  P 4 ) (F 2  V x ) ^ . g ,
heat taken in 778 (x^Li + h — fe) ?
where the denominator represents the amount of heat taken in,
with the feedwater at temperature of exhaust, t 2 . In actual
practice the efficiency of engines using steam without expansion
is about 0.06 to 0.07, when the temperature of condensation is
about ioo° F. When steam is used in an engine without ex
pansion and also without the use of a condenser the value of
this efficiency is still lower. It will be observed that under the
most favorable conditions obtainable the efficiency of an engine
without expansion cannot be made under normal conditions to
exceed about 7 per cent.
In the actual Newcomen steam engines the efficiency was very
much lower than any of the values given because at every stroke
of the piston a very much larger amount of steam had to be
taken in than that corresponding to the volume swept through
by the piston on account of a considerable quantity of steam
condensing on the walls of the cylinder.
Adiabatic Expansion and Available Energy. A practical ex
ample as to how the temperatureentropy diagram can be used
to show how much work can be obtained by a theoretically per
fect engine from the adiabatic expansion of a pound of steam
will now be given. When steam expands adiabatically — with
out a gain or loss of heat by conduction — its temperature falls.
Remembering that areas in the temperatureentropy diagram
represent quantities of heat and that in this expansion there is
no exchange of heat, it is obvious that the area under a curve of
adiabatic expansion must be zero ; this condition can be satisfied
only by a vertical line which is a line of constant entropy.
The work done during an adiabatic process, while it cannot be
obtained from a "heat diagram," can very readily be determined
from the area under the adiabatic curve of a pressurevolume
diagram, or better still by the use of steam tables as follows: In
an adiabatic expansion the amount of work done is the mechani
cal equivalent of the loss in internal energy as explained in Chap
ter III. Therefore, it is only necessary to determine the internal
ADIABATIC EXPANSION AND AVAILABLE ENERGY 129
energy of the steam at the beginning and end of the adiabatic
expansion.
h = h x + Xipi, (182)
I 2 = h 2 + x 2 p 2 . (183)
Work during adiabatic expansion = loss in internal energy
W = (hi + Xipi — Jh — X2P2) 778 (footpounds). (184)
Fig 37 is a temperatureentropy diagram representing dry
saturated steam which is expanded adiabatically from an ini
tial temperature Z\ (corresponding to a pressure Pi) to a lower
final temperature T 2 (corresponding to a pressure P 2 ), The
C / T, P, \
E
JlBB
"J
^^^^^^^^^
G \
g
1
F
F'
&
Entropy a r 2
Fig. 37. — Temperatureentropy Diagram for Dry Saturated Steam
Expanded Adiabatically.
initial and final conditions of total heat (H) and entropy (<p)
are represented by the same subscripts 1 and 2. The available
energy or the work that can be done by a perfect engine under
these conditions is the area NCEG. It is now desired to obtain
a simple equation expressing this available energy E a in terms
of total heat, absolute temperature and entropy.
E x = area OBNCEF,
H 2 = area OBNG / F / ,
E a = area (OBNCEF + FGG'F')  OBNG'F',
E a = H 1 H 2 + (c P 2 <l>i) 2V* (185)
* It should be observed that this form is for the case where the steam is
initially dry and saturated. For the case of superheated steam a slightly differ
ent form is required.
130 CYCLES OF HEAT ENGINES USING VAPORS
An application of this equation will be made to determine the
heat energy available from the adiabatic expansion of a pound
of dry saturated steam from an initial pressure of 165 pounds
per square inch absolute to a final pressure of 15 pounds per
square inch absolute.
Example. Pi = 165 7\ = 826 degrees, from steam tables.
P 2 = IS T 2 = 673.0
#1 = 1195.0B.tu. "
#2 = 1150.7 B.t.u. "
01 = I56I5
<h = 1.7549
Substituting these values in equation (185), we have
E a = 1195.0  1150.7 + (i.7549  15615) 673 = 174.5 B.t.u.
per pound of steam.
Now if in a suitable piece of apparatus like a steam turbine
nozzle, all this energy that is theoretically available could be
changed into velocity, then we have by the wellknown formula
in mechanics, for unit mass,*
V 2
— = E a (footpounds) = E a (B.t.u.) X 778,
2 g
V = V778 X 2 gE a = 223.8 VE a , (l86)
where V is the velocity of the jet and g is the acceleration due to
gravity (32.2), both in feet per second.
Solving then for the theoretical velocity obtainable from the
available energy we obtain the following:
V = 223.8 V174.5 = 223.8 X 13.22 = 2956 feet per second.
The important condition assumed as the basis for determining
equation (185), that the steam is initially dry and saturated, must
not be overlooked in its application. There are, therefore, two
other cases to be considered:
(1) When the steam is initially wet,
(2) When the steam is initially superheated.
* See Church's Mechanics of Engineering, page 672, or Jameson's Applied Me
chanics and Mechanical Engineering, vol. I, page 47.
AVAILABLE ENERGY OF WET STEAM
131
Available Energy of Wet Steam. The case of initially wet
steam is easily treated in the same way as dry and saturated
steam. Fig. 38 is an example of the case in hand. At the ini
tial pressure P\ the total heat of a pound of wet steam (hi + X1L1)
is represented in this diagram by the area OBNCE^F". The
initial quality of the steam (xi) is represented by the ratio of the
CE"
The available energy from adiabatic expansion from
lines
CE
the initial temperature Ti (corresponding to the pressure Pi) to
Entropy (p x ^
Fig. 38. — Temperatureentropy Diagram of Wet Steam Expanded Adiabatically.
the final temperature T 2 (corresponding to the pressure P 2 ) is the
area NCE"G". If we call this available energy E aw , we have by
manipulation of the areas,
Eaw = area OBNCEF + FGG'F'  OBNG'F'  G"E"EG,
E aw = H 1 H 2 + (<p 2  fa) T 2  (0! 0,) {Ti  T 2 ) * (187)
or,
E aw = H l H 2 + (0 2  0i) T 2  ^ (1  Xi) (Ti  T 2 ). (188)
J 1
The velocity corresponding to this energy is found by substi
tution in equation (186), just as for the case when the steam was
initially dry and saturated.
■t 1
<t>X = Xi 7jT + 01, 01 — #6 = J (i — #l).
1 1 ll
132
CYCLES OF HEAT ENGINES USING VAPORS
Example. Calculations for the velocity resulting from adia
batic expansion for the same conditions given in the preceding
example, except that the steam is initially 5 per cent wet, are
given below.
Pi = 165 lbs. absolute. Ti = 826 degrees from tables
P 2 = 15 lbs. absolute. T 2 = 673.0 degrees
Hi = 1195.0 B.t.u.
H 2 = 1150.7 "
0i = 1.5615
02 = 17549
Li = 856.8 B.t.u.
Xi = 1. 00 — 0.05 = 0.95
E c
(1195.0  1150.7) + (i.7549  15615) 673 
856.8
826
X (i  0.95) (826  673) = 44.3 + 130.2  7.93
Eaw = 166.5 B.t.u. per pound of wet steam
V = 223.8 VeZ = 223.8 x V166.5
= 223.8 X 12.9 = 2886 feet per second.
This result can be checked very quickly by the "total heat
entropy" or "Mollier" diagram (Appendix). The intersection
of the 0.95 quality line and 165 pounds pressure line is found to
lie on the n 52 B.t.u. total heat line. Since the expansion is
adiabatic, the entropy remains constant. Therefore, following
the vertical or constant entropy line (entropy = about 1.507)
down to its intersection with the 1 5 pounds pressure line, we find
that the total heat at the end of adiabatic expansion is 985 B.t.u
and
1 152 — 985 = 167 B.t.u. available energy, as above.
If the steam were initially superheated, the available energy
during adiabatic expansion could be obtained in the same way by
means of the diagram.
Available Energy of Superheated Steam. The amount of
energy that becomes available in the adiabatic expansion of
superheated steam is very easily expressed with the help of Fig.
39. Two conditions after expansion must be considered:
AVAILABLE ENERGY OF SUPERHEATED STEAM
1 S3
(i) When the steam in the final condition is superheated,
(2) When the steam in the final condition is wet (or dry
saturated).
Using Fig. 39 with the notation as before except E as is the avail
able energy from the adiabatic expansion of steam initially super
heated in B.t.u. per pound, S and H s are respectively the total
entropy and the total heat of the superheated steam at the
T 3 H3
l*
Ti
Entropy — <j>
Fig. 39. — Temperatureentropy Diagram for Superheated Sxeam.
initial condition, then from the diagram, when the steam is
wet at the final condition, •
E fls = H S H 2 + (0 2  <j> s ) T 2 . (189)
When the steam is superheated at the final condition,
E as = H s  H 2 ;  (0,  2 O TV. (190)
It will be observed that these equations (189) and (190) are the
same in form as (184), and that equation (190) differs only in hav
134 CYCLES OF HEAT ENGINES USING VAPORS
ing the terms H s and fa in the place of Hi and fa. In other
words equation (184) can be used for superheated steam if the
total heat and entropy are read from the steam tables for the
required degrees of initial superheat.
The following examples illustrate the simplicity of calcula
tions with these equations:
Example 1. Steam at 150 pounds per square inch absolute
pressure and 300 F. superheat is expanded adiabatically to 1
pound per square inch absolute pressure. How much energy in
B.t.u. per pound is made available for doing work?
Solution. H s = 1348.8 B.t.u. per pound,
H 2 = 1 103.6 "
fa = 1.980,
4>> = i73 2 >
. T 2 = 5596° F,
Eas = I348.8  IIO3.6 + (1.980  I.732) 559.6
= 383.9 B.t.u. per pound.
The result above may be checked with the total heatentropy
chart (Appendix) and obtain thus (1349 — 967) or 382 B.t.u.
per pound.
Example 2. Data the same as in preceding example except
that the final pressure is now 35 pounds per square inch absolute.
(Final condition of steam is superheated.) Calculate Eas.
Solution. H s = 1348.8 B.t.u. per pound,
HJ = 1 166.8 " " "
fa = 1.732,
fa' = 1.6868,
TV = 718.9° F.,
Eas = 1348.8  1166.8  (1.7320  1.6868) 718.9
= 149.5 B.t.u. per pound.
Application of Temperatureentropy Diagram to Analysis of
Steam Engine. The working conditions of a steam engine, as
stated before, can be shown not only by the indicator card, but
also by the employment of what is known as a "temperature
APPLICATION OF TEMPERATUREENTROPY DIAGRAM 135
entropy" diagram. This diagram represents graphically the
amount of heat actually transformed into work, and in addition
the distribution of losses, in the steam engine.
For illustration, a card was taken from a Corliss steam engine
having a cylinder volume of 1.325 cubic feet, with a clearance
volume of 7.74 per cent, or 0.103 cubic feet; the weight of steam
in pounds per stroke (cylinder feed plus clearance) was 0.14664
pounds. Barometer registered atmospheric condition as 14.5
pounds per square inch. The scale of the indicator spring used
in getting the card was 80 pounds to the inch. Steam chest
pressure was taken as 153 pounds per square inch (absolute),
and a calorimeter determination showed the steam to be dry
and saturated.
The preliminary work in transferring the indicator card to a
T<f> diagram, consists first in preparing the indicator card as
follows: It was divided into horizontal strips at pressure intervals
of 10 pounds with the absolute zero line taken as a reference;
this line was laid off 14.5 pounds below atmospheric conditions.
(See Fig. 40.) For reference, the saturation curve was drawn.
Knowing the weight of steam consumed per stroke and the
specific volume of the steam (from the Steam Tables), for
various pressures taken from the card, the corresponding actual
volumes could be obtained; this operation is, merely, weight of
steam per stroke multiplied by specific volume for some pressure
(0.14664 X column 5 in the table below), the resulting value be
ing the volume in cubic feet for that condition. These pressures
and volumes were plotted on the card and the points joined, re
sulting in the saturation curve, 2 // 6"8 // o/ / .
The next step consisted in constructing the " transformation
table " with the columns headed as shown. All the condensing
and evaporation processes are assumed to take place in the cylin
der and the Tcf> diagram is then worked up for a total weight of
one pound of steam as is customary. Column 1 shows the re
spective point numbers that were noted on the card; column 2,
the absolute pressures for such points; column 3, the corre
sponding temperatures for such pressures; column 4, the vol
136
CYCLES OF HEAT ENGINES USING VAPORS
ENTROPY DIAGRAM
1.0
Entropy
Fig. 40. — Temperatureentropy Diagram of Actual Steam Engine Indicator
Diagram.
APPLICATION OF TEMPERATUREENTROPY DIAGRAM 137
ume in cubic feet up to the particular point measured from the
reference line of volumes; column 5, the specific volume of a
pound of dry and saturated steam at the particular pressure
(Steam Tables); column 6, the volume of actual steam per
pound, obtained by dividing the volumes in column 4 by 0.14664
pound (total weight of steam in cylinder per stroke) ; column 7,
the dryness fraction "x," found by dividing column 6 by column
5 ; column 8 is the entropy of evaporation ( — J f or particular con
ditions (Steam Tables); column 9 is the product of column 7
and column 8; column 10 is the entropy of the liquid at various
conditions as found in the Steam Tables; column 11 is the sum
of column 9 and column 10, giving the total entropy.
TRANSFORMATION TABLE
I
2
3
4
5
6
7
8
9
IO
11
u
0>
.a
6
a
u
<a
a,
tA w
,0.0
— ' a)
w .4.
<l> xn
u
14
O.S
•^
s.s
>8
in
."gto
"c3
CJ» ' _
C8 «...
O ft** 1
8 6 3
O w
>
CO
C O
.2
P. njH
O U \
'B
a"—
u
a
a
2
H
1
2
5
9
12
is
18
20
145
140
I20
48
20
8
8
3°
356
353
34i
279
228
183
183
250
0.1050
0.2250
0.4230
0.8500
1 .4225
0.9758
0.3500
0.1825
3
3
3
8
20
47
47
13
11
22
73
84
08
27
27
74
0.716
1536
2 .89O
58io
9.700
6.660
2.390
1.245
0.2302
0.4770
0.7740
0.6580
0.4830
0.1410
. 0506
0.0907
I .0612
I .0675
10954
12536
13965
15380
15380
I33H
O . 2440
O . 5090
0.8475
O.825O
O.674O
0.2l68
O.O778
O . I 208
0.5107
0.5072
0.4919
0.4077
03355
0.2673
0.2637
0.3680
0.7547
1 .0162
13394
1.2327
1.0095
0.4841
0345I
0.4888
Above table is employed for transferring the PV diagram to the T<t> diagram.
After this table was completed, columns 3 and 1 1 were plotted
(Fig. 40). Convenient scales were selected, the ordinates as tem
peratures and the abscissas as entropies. The various points,
properly designated, were connected as shown on the T<f> dia
gram, the closed diagram resulting. This area shows the amount
of heat actually transformed into work. This diagram is the
actual temperatureentropy diagram for the card taken and may
138 CYCLES OF HEAT ENGINES USING VAPORS
•
be superimposed upon the Rankine cycle diagram in order to
determine the amount and distribution of heat losses.
The water line, AA', and the dry steam line, CC , were
drawn directly by the aid of Steam Table data, i.e., the entropy
of the liquid and the entropy of the steam taken at various
temperatures, and plotted accordingly.
Before the figure could be studied to any extent, the theoreti
cal (Rankine) diagram had to be plotted, assuming that the steam
reaches cutoff under steamchest conditions; that it then ex
pands adiabatically down to back pressure and finally exhausts
at constant pressure to the end of the stroke without compres
sion. This diagram is marked, ACEH, on the T4> plane.
The steamchest pressure of 153 pounds per square inch absolute
fixes the point, C, when the temperature line cuts the steam line,
CC . The rest of the cycle is selfevident.
Referring to the T<f> diagram, Fig. 40, HACE is the Ran
kine cycle with no clearance for one pound of working fluid.
The amount of heat supplied is shown by the area, MiHACN,
and of this quantity, the area M\HEN * would be lost in
the exhaust while the remainder, HACE, would go into
work. This is theoretical, but in practice there are losses, and
for that reason, the Rankine cycle is used merely for comparison
with the actual card as taken from a test. The enclosed irregu
lar area, 123 • • • 222 3> * s the amount of heat going into
actual work. By observation, it is evident that a big area re
mains; this must represent losses of some sort or other. That
quantity of work represented by the area, i55 / i / , is lost on
account of wiredrawing; the area $'CDF shows a loss due
to initial condensation; the loss due to early release is shown
by the area F1214F' for the real card, and by DGE for the
modified Rankine cycle (such a loss, in other words, is due to
incomplete expansion); that quantity represented by 22Bi f i
is lost on account of incomplete compression, and HABiS is
* The areas Mi HACN and Mi HEN should have added to them, the rectangu
lar area between MiN, and the absolute zero line, which is not shown on the
diagram (Fig. 40).
APPLICATION OF TEMPERATUREENTROPY DIAGRAM 139
1.6 2.0
Entropy
Fig. 41. — Temperature (absolute) entropy Diagram of Actual Steam Engine
Indicator Diagram.
140 CYCLES OF HEAT ENGINES USING VAPORS
the loss due to clearance. The expansion line from 5 on to 9 in
dicates that there is a loss of heat to the cylinder walls, causing
a decrease of entropy; from 9 on to 10, reevaporation is taking
place (showing a gain of entropy).
All of the heat losses are not necessarily due to the transfer
of heat to or from the steam, as there may be some loss of steam
due to leakage. In general, however, the T<f> diagram is satis
factory in showing heat losses.
Fig. 41 was constructed for the purpose of showing how the
actual thermal efficiency and the theoretical thermal efficiency
(based on the Rankine cycle) can be obtained from the T<f>
diagram. The letters in Fig. 40 refer to the same points as
in Fig. 41, the only difference between the two diagrams being
the addition of the absolute zero temperature line to Fig. 41.
Thermal efficiency =
Actual thermal efficiency =
Work done
Heat added
Shaded area W
M^HACN''
= —7^ — . (Planimeter)
20.00 sq. in.
= 0.096, or 9.6 per cent.*
Rankine cycle efficiency (theoretical thermal efficiency)
HACE
" Mi'HACN''
= ~.* " (by planimeter)
20.00
= 0.203, or 20.3 per cent.
Combined Indicator Card of Compound Engine. The method
of constructing indicator diagrams to a common scale of vol
* This value may be regarded as the actual thermal efficiency for one stroke.
inasmuch as in the succeeding strokes the "cushion" steam will be used over and
over again and hence will not constitute a heat loss. The actual thermal efficiency
W
of a steam engine under running conditions would then be given by , ^rff
where 18 BCE is a Rankine cycle with clearance and complete compression.
COMBINED INDICATOR CARD OF COMPOUND ENGINE 141
ume and pressure shows where the losses peculiar to a compound
steam engine occur, and, to the same scale, the relative work
areas.
As the first step divide the length of the original indicator dia
grams into any number of equal parts (Fig. 42), erecting per
pendiculars at the points of division.
In constructing a combined card, select a scale of absolute
pressure for the ordinates and a scale of volumes in cubic feet
145.6*
S7.ir
Zero Line
Fig. 42. — High and Low Pressure Indicator Diagram of Compound
Steam Engine.
for the abscissas. To the scale adopted draw in the atmos
pheric pressure (" at.") line, see Fig. 43.
Lay off the lowpressure clearance volume on the xaxis to
the scale selected. In like manner lay off the piston displace
ment of the lowpressure cylinder and divide this length into the
same number of equal parts as the original indicator diagrams
were divided. From the original lowpressure card (Fig. 42), de
termine the pressures at the points of intersection of the perpen
diculars erected above the line of zero pressure, taking care
that the proper indicatorspring scale is used. Lay off these
pressures along the ordinates (Fig. 43), connect the points and
the result will be the lowpressure diagram transferred to the
new volume and pressure scales.
142 CYCLES OF HEAT ENGINES USING VAPORS
The highpressure diagram is transferred to the new volume
and pressure scale by exactly the same means as described for
the lowpressure diagram.
The saturation curve is next drawn. This curve represents
the curve of expansion which would be obtained if all the steam
in the cylinder was dry and saturated. It is very probable
that the saturation curve for each cylinder will not be continuous
since the weight of the cushion steam in the lowpressure cylin
der is usually not the same as that in the highpressure cylinder.
The saturation curve would be continuous for the two cards only
when the weight of cushion steam in the highpressure and low
pressure cylinders is the same (assuming no leakage or other
losses).
On the assumption that the steam caught in the clearance
spaces at the beginning of compression is dry and saturated, the
weight of the cushion steam can be calculated from the pressure
at the beginning or end of compression, the corresponding cylin
der volumes and the specific volumes corresponding to the pres
sure (as obtained from steam tables).
The total weight of steam in the cylinder is the weight of steam
taken into the engine per stroke plus the weight of steam caught
in the clearance space (cushion steam). The saturation curves
may now be drawn by plotting the volumes which the total
weight of steam will occupy at different pressures, assuming it
to be dry throughout the stroke.
The quality curve (Fig. 43) shows the condition of the steam
as the expansion goes on. At any given absolute pressure, the
volume up to the expansion line shows the volume of the wet
vapor, while the volume up to the saturation curve shows the
volume that the weight of the wet vapor would have if it were
dry. Thus at any given absolute pressure, the ratio of the vol
ume of the wet vapor (as given by the expansion line of the
indicator card) to the total volume of the dry vapor (as obtained
from the saturation curve) is the measure of the quality of the
steam.
Showing the quality by the use of the figure, we have Vol.
COMBINED INDICATOR CARD OF COMPOUND ENGINE 143
160
155
150
145
140
135
130
125
120
115
110
105
100
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Fig. 43. — Combined Indicator Diagrams for Compound Engine.
144 CYCLES OF HEAT ENGINES USING VAPORS
ab v Vol. ac = quality. By laying off this ratio from a hori
zontal line to any scale desired as shown, the quality curve may
be constructed.
Hirn's Analysis. In the study of steamengine performance
the action of the steam in and the influence of the walls of the
cylinder become a matter of considerable importance. The
amount of heat lost, restored and transformed into work as a
result of variation in the condition of the steam throughout the
enginecycle can be determined either by means of the entropy
temperature diagram, or by calorimetric method based on Hirn's
theory.
The calorimetric method, or more popularly called Hirn'r,
Analysis, was developed by Professor V. DevelshauversDevy of
Liege. (See Table of the Properties of Steam by V. Devels
hauversDevy, Trans. Am. Soc. Mech. Eng., Vol. XI; also Pea
body's Thermodynamics.)
In order to apply Hirn's Analysis to engine testing besides the
usual readings the following items must be determined :
i. Absolute pressures at cutoff, release and compression from
indicator cards,
2. Per cent of stroke at cutoff, release and compression from
indicator cards,
3. Weight of steam per stroke determined by weighing the
condensed steam or boiler feed water and computing the steam
used per stroke,
4. Weight of the cooling water and its temperature at inlet
and outlet from condenser,
5. Temperature of the condensed steam.
Let w = pounds of steam supplied to the cylinder per stroke
at pressure p and quality x.
Then the amount of heat brought in by the steam into the
cylinder per stroke,
Q = w (h + xL), (191)
where L = heat of vaporization and h = heat of the liquid
above the freezing point.
HIRN'S ANALYSIS
145
At the end of compression a certain amount of steam is left
in the clearance space and this steam mingles with the w pounds
of steam at admission. Calling the weight of the steam caught
in the clearance space Wi, the total amount of steam at admission
is w f w\.
According to thermodynamic laws the addition of heat to a
substance produces in that substance internal and external
changes, the internal changes being volume and temperature
variations which are called intrinsic energy changes, while the
Volume
Fig. 44. — The Steam Engine Cycle.
external changes are changes in external potential energy or work
done. Thus the steam brought into the cylinder per stroke on
account of its available heat produces intrinsic energy changes
and is capable of doing external work in overcoming resistance.
On the other hand the steam W\ caught in the clearance space at
compression is able to produce only internal or intrinsic energy
changes. The intrinsic energy changes can be calculated knowing
the heat equivalent of internal work or internal latent heat (p)
and the heat of the liquid Qi) . The external work can be deter
mined from the indicator cards.
Referring to Fig. 44 and calling the heat absorbed by the
cylinder walls during admission and expansion Q a and Q b ', that
restored during exhaust and compression Q c and Q d ] the intrinsic
146 CYCLES OF HEAT ENGINES USING VAPORS
energy at the points of admission, cutoff, release, and compres
sion /1, I 2 , Iz, I *', the external work in heat units AW a , AW b ,
AW C , AWd, during the events of the stroke, where A = t™;
also if the temperature of the condensed steam is t s and that of
the cooling water U and U at inlet and outlet, the total weight of
cooling water used being G, we have the following relations:
Qa = Q + h ~ h ~ AWa. (192)
Q b = hh AW b . (193)
Q c = I3 — I4 — wh s — G (h — hi) — AW C . (194)
Q d = hh + AW d . (195)
In equations (192), (193), (194), and (195) Q can be determined
from Q = w (h + xL). The values of G can be determined by
weighing cooling water; h , ht, and h s by taking the temperature
t 0i U, and t s by thermometers and finding h , hi, and h s from steam
tables. The intrinsic energy at the events of the stroke are calcu
lated by means of the following equations:
h = Wi (h + tfipi), (196)
h = (wi J w) (h 2 + X2P2), (197)
h = (wi + w) (h z + ^sPs), (198)
7 4 = Wi (h + ^4p4), (199)
in which
W\ — weight of steam caught in clearance space.
w — weight of steam brought into the cylinder per
stroke.
hi, 1h, h z , hi = heat of the liquid at events of stroke.
Pi, p2, P3, pi = internal latent heats at the events of stroke.
%i, x 2 , x 3 , x± = quality of steam at events of stroke.
From the above it is evident that Xi, x 2 , x 3 , #4, and w are un
known, these values being determined by the following procedure:
The volume of w pounds of steam in cubic feet is
V = w (xu + 0). (200)
x = quality of steam, u = the increase in volume produced by
the vaporization of one pound of water of volume a into dry
CLAYTON'S ANALYSIS OF CYLINDER PERFORMANCE 147
steam of volume S, or u = S — a. Calling the volume of the
steam caught in the clearance space Vi and that at cutoff, release
and compression V 2 , V3, F 4 ,
Vi = Wi(xiUi + a). (201)
Vi + V 2 = (w + Wi) (x 2 u 2 + a). (202)
Vi + V 3 = (w + Wi) (X3U3 + a). (203)
Vi + Va = W\ (X4U4 + <r). (204)
In the above equations the absolute pressures and volumes can
be determined from the indicator cards, provided the dimensions
of the engine and the clearance are known and it is assumed
# 4 is unity. This assumption can be made without much error,
as the steam at compression is very nearly dry. If x± is assumed
equal to 1, then, from equation (204),
V 1 + F 4 f ,
w l = (205)
U\ + cr
The intrinsic energy at the events of the stroke is computed
after finding x h x 2 , and x 3 from equations (201), (202), and (203)
by the aid of equation (205).*
Clayton's Analysis of Cylinder Performance. The indicator
diagram has been used in determining the economy of the steam
engine although inaccuracies enter because of cylinder condensa
tion. The quantity of steam admitted to the engine per cycle
can theoretically be determined from the difference between the
weight of steam accounted for at the point of cutoff and point of
compression. Little error is introduced in determining the
weight of steam present in the cylinder at the point of compres
sion, for the quality of the steam at that point can be fairly accu
rately estimated. To determine the weight of steam present at
the point of cutoff the quality of the steam must be known. The
quality of the steam at the point of cutoff is not the same as that
at admission because of the condensation within the cylinder.
Thus the study of cylinder performance from the indicator dia
gram is made inaccurate.
* For log form of Hirn's Analysis see Experimental Engineering by Carpenter
and Diederichs, Seventh Edition, pages 799806.
148 CYCLES OF HEAT ENGINES USING VAPORS
Clayton's analysis was made to determine the relation be
tween the quality of the steam at cutoff and other variables.
From the results of a careful study of the forms of the expansion
and compression curves which occur in indicator diagrams, it has
been discovered that the value of n for the expansion curve bears
a definite relation to the proportion of the total weight of steam
mixture which was present at the point of cutoff.
To facilitate the analysis, the indicator diagram is transferred
to logarithmic coordinates. The equation of the poly tropic
curve, PV n = C, becomes a straight line when plotted upon
logarithmic crosssection paper, the value of n becoming the slope
of the curve. From the new diagram, the slope of the expansion
and compression curves may be easily determined. The quality
of the steam at cutoff may then be estimated, knowing the speed
of the engine, the quality of the steam at admission, the pressure
of the steam and other values that affect the slope of the curve.*
PROBLEMS
1. Assume that 1 lb. of steam of a pressure of 160 lbs. per sq. in. abso
lute and a quality of 0.95 performs an ideal Rankine cycle, being exhausted
at a pressure of 5 lbs. per sq. in. absolute. Compute the quality of the steam
exhausted, the efficiency of the cycle, and the final volume.
2. What is the work of an ideal Rankine cycle if the steam initially at
200 lbs. per sq. in. absolute pressure, superheated 200 F., goes through such
a cycle with a back pressure of 1 lb. per sq. in. absolute?
3. One pound of steam at a pressure of 100 lbs. per sq. in. absolute
with a quality of 0.90 performs an ideal Rankine cycle exhausting at a back
pressure of 2 lbs. per sq. in. absolute. ' What is the net work and efficiency
of the cycle?
4. Two pounds of steam at a pressure of 125 lbs. per sq. in. absolute
and a volume of 8.34 cu. ft. perform an ideal Rankine cycle. The exhaust
pressure is 25 lbs. per sq. in. absolute. What is the net work and efficiency
of the cycle?
5. One pound of steam at a pressure of 160 lbs. per sq. in. absolute and
a quality of 0,95 passes through a modified Rankine cycle. The terminal
* For details concerning Clayton's "Analysis of the Cylinder Performance
of Reciprocating Engines," see Bulletin No. 58 of the Engineering Experiment
Station of the University of Illinois.
PROBLEMS 149
pressure is 11 lbs. per sq. in. absolute and the exhaust pressure 5 lbs. per sq.
in. absolute. What is the efficiency of the cycle?
6. One pound of steam at a pressure of 200 lbs. per sq. in. absolute and
200 F. superheat passes through a modified Rankine cycle. The terminal
pressure is 15 lbs. per sq. in. absolute and the exhaust pressure 10 lbs. per
sq. in. absolute. What is the net work and efficiency of the cycle?
7. Two pounds of steam at a pressure of 125 lbs. per sq. in. absolute and
volume of 8.34 cu. ft. pass through a modified Rankine cycle. The terminal
pressure is 20 lbs. per sq. in. absolute and the exhaust pressure 15 lbs. per
sq. in. absolute. What is the efficiency, net work, and heat added to the
cycle?
8. A noncondensing steam engine receives steam of 0.95 quality and
100 lbs. per sq. in. absolute pressure. The exhaust pressure is 15 lbs. per
sq. in. absolute and the steam consumption of the engine is 30 lbs. per
indicated horse power per hour. What is the thermal efficiency of the
engine? What is the efficiency of the ideal Rankine cycle operating under
the same conditions? What is the efficiency of the Carnot cycle operating
between the same temperature limits?
Throat or smallest
section of nozzle
CHAPTER IX
FLOW OF FLUIDS
Flow through a Nozzle or Orifice. Thermodynamic prob
lems embrace the measurement of the flow of air or of a mix
ture of a liquid and vapor through a nozzle or orifice. In the
nozzle shown in Fig. 45, let A and B
be two sections through which the sub
stance passes. At A let a pressure of
Pi be maintained and at B a pressure
of P 2 . To maintain the constant pres
sure at A of Pi let more substance be
Fig. 45.   Typical Nozzle added, while at B allow enough of the
for Expanding Gases and Va substance to be discharged so that the
pors * constant pressure of P 2 is maintained.
The quantity of energy and the mass passing into the section A
must be accounted for at section B and the relation of these
quantities will determine the change of the velocity of the sub
stance.
After uniform conditions have been established in the nozzle,
the same mass entering at A must be discharged at B during
the same time. Thus any mass may be considered as a work
ing basis, but as a rule one pound of the substance is used. All
formulas refer to one pound, unless another mass is definitely
stated.
The same quantity of energy discharged at B must enter at
A unless heat is added to or taken from the substance between
the sections A and B. Thus the general formula is derived:
Energy carried by substance at B = energy carried by sub
stance at A + energy added between the sections A and B.
The energy carried by the substance at the entering or dis
charge end of the nozzle is made up of three quantities: (1) the
150
FLOW THROUGH A NOZZLE OR ORIFICE 151
amount of work necessary to maintain constant pressure at each
end of the nozzle; (2) the internal energy of the substance,
(3) the kinetic energy stored in the substance because of the
velocity which it has when passing the section.
The amount of work necessary to maintain a constant pres
sure (pounds per square foot) of Pi at A or of P 2 at B is — ^ or
778
p „.
—~, where i\ and % are the volumes of one pound of the sub
778
stance at A and B respectively.
The internal or " intrinsic " energy in B.t.u. per pound (I H )
of the substance passing A or B, calculating from 32 F. is for:
Air or similar gases, \ ? (206)
5 ' 778X040
Liquid, h t (207)
/ 72 \
Liquid and Vapor, h + x ( L )> (208)
H V 2 g X 778/
/ V 2 \
Superheated Vapor, h + \L )
F V 2 g X 778/
+c y (r SUP r sa t.) jP(w 7 z;sat) , (209)
778
V 2
where is the kinetic energy in B.t.u. per pound of the
2 g X 778
substance as it passes a section, V = velocity in feet per second
and g = 32.2 feet.
If Q represents in B.t.u. per pound the heat units added to the
substance between the sections A and B (Fig. 45), the energy
equation for air and similar gases can be found by equating the
total heat energy put in at A plus the energy added between
A and B to the energy discharged at B. This general formula
reduces to
F 2 2  TV = 2 g [j^ (P 1 v 1  P 2 v 2 ) + 778 <2~j
or = 2 g X 778 [C 9 (Xi  T 2 ) + Q]. (210)
* See equation (61).
152 FLOW OF FLUIDS
This thermodynamic equation is the usual form for the flow
of air or similar gases.
The energy equation for superheated steam can be derived
by the use of the same general formula stated above, on the
assumption that the substance is superheated at A and B. This
formula reduces to
= [h + L\ + C p (r si ,p. — Taa,t)i]
2 g X 778
— [h 2 + l 2 + c P (r SUP . — r sa t.) 2 ] + Q, (211)
or V 2 2 — V1 2 = 2 g X 778 [Tabular heat content* — Tabu
lar heat content^ + Q], (212)
From a thermodynamic standpoint, the relation between the
initial and final condition is that of adiabatic expansion when all
the heat which disappears as such is used in changing the veloc
ity, provided the nozzle is properly shaped and Q is zero. The
diagram in Fig. 46 represents this condition of affairs on a tem
peratureentropy diagram for air and similar substances. The
area acdf is C v (T 2 — 492), and the area abef is C p ( Ji — 492).
The quantity of heat energy changed into kinetic energy is there
fore the area bcde and is the difference between the internal
energy in the substance at the beginning and end of the opera
tion, together with the excess of work done to maintain the pres
sure of P± at A over the pressure of P 2 at B. The line cd would
incline to the right if heat were added in the nozzle, since the
effect would be to increase the velocity or increase the area cdeb.
The line cd would incline to the left from c if heat were lost,
since the area cdeb would decrease.
In Fig. 47 the diagram represents the conditions for super
heated vapor. The area aa'cdf represents the heat required to
raise the substance from a liquid at 32 F. to superheated vapor
at the temperature of Tisup. The area ab'bdf is the heat con
tent at b, the final condition. The area a'cbb' represents, there
fore, the heat available for increasing the velocity. The areas
* Tabular heat content means the total heat of superheated steam as read
from tables of the properties of superheated steam.
FLOW THROUGH A NOZZLE OR ORIFICE
153
representing the heat available for increasing the velocity in Fig.
46 and Fig. 47 are shown by the crosshatched area in Fig. 48
and are really the representation of the work done (theoretically)
in an engine giving such an indicator diagram.
Evidently the greater the drop
in pressure, the greater will be the
crosshatched area in Fig. 48 and
the greater will be the velocity,
T 2
c n
T,
g\
V
a/
492
T. Ent. Diagram
i
1
/
la
d\\
t^up. OJ
Tjsat. a' 1
T 2 sup. / \
6
T2 sat. / \
r
U T.Ent.Diagram
d
Fig. 46. — Temperature
entropy Diagram of Heat
Available in Air.
Fig. 47. — Temperatureentropy Dia
gram of the Heat Available in Superheated
Steam for Increasing Velocity.
regardless of the substance. The line ab in Fig. 49 represents
p
the velocity curve with V as ordinates and ^ as abscissas, but
" 1
since with any substance ex ! i
panding the weight of a cubic
foot decreases as the pressure
drops, the line cd will repre
sent to some scale not here,
determined the weight of a
cubic foot at any discharge
pressure
Fig. 48. — Heat (Work) Available for
Increasing Velocity.
Since the product of the area through which the discharge
takes place, the velocity and the weight per cubic foot of the
154
FLOW OF FLUIDS
substance is equivalent to the weight in pounds of the substance
discharged per second, the product of the ordinates at any point
of the curves ab and cd is proportional to the weight discharged
from a pressure of P\ to a condition where the pressure is P 2 .
The line ceb represents this product. Evidently there is some
low pressure into which the weight discharged per square foot
will be a maximum, and this will be the pressure corresponding
to the high point e on the curve.
a
V
e
\f
v in.p
^
/4
d
\ c
&
^
f
$?
l/c
b
.2
.5
.6 .7
.9 1.0
Fig. 49. — Illustrative Curves of Weight, Discharge and Velocity.
Weight per Cubic Foot. From the formula for adiabatic ex
pansion the weight per cubic foot can be obtained if the sub
stance is similar to air.* The general formula (applied to air) is
(213)
(214)
* The exponent in the formula is the ratio of the specific heats (of air in this
case).
P^l40 B
P2^ 40 ,
q be reduced to
1
1 P 2 14 X Pi, t
Pi 1  4 X RTi
or
, p 0.286 p 0.714
v 2 RTi
FLOW OF AIR THROUGH AN ORIFICE 155
which is the weight in pounds per cubic foot of discharge. Pres
sures are in pounds per square foot. If the supply to the nozzle
is from a large reservoir so that V± can be taken as zero then
the discharge velocity is
T 0.4
v,  V/Vx^Wxg)
v, = 109.6 \/ r, [1  (0' 286 ] (21s)
All quantities on the righthand side of this equation must be
obtained from the data of tests. • Weight in pounds discharged
through the area A (in square feet) is
P 0.286 P„0714
w = A x Fl Jj xi
Kl 1
o 9 .6v/ r,[i  (0""} (216)
Maximum Discharge. This weight is a maximum when
—  = o or when P 2 = 0.525 Pi. The maximum quantity of air
aP 2
will be discharged when the low pressure is 52.5 per cent of the
high pressure.
Shape of Nozzle. See page 167, on the flow of steam.
Flow of Air through an Orifice. Air under comparatively high
pressure is usually measured in practice by means of pressure
and temperature observations made on the two sides of a sharp
edged orifice in a diaphragm. The method requires the use of
two pressure gages on opposite sides of the orifice and a ther
mometer for obtaining the temperature ti at the initial or higher
pressure pi. The flow of air w, in pounds per second, may then
be calculated by Fliegner's formulas :
w = 0.530 X f X a— t^z when pi is greater than 2 p 2 , (217)
vTi
w = 1.060 Xf X a 1 /P2 (Pi ~ P2) w hen p x is less than 2 p 2 , (218)
V Ti
156 FLOW OF FLUIDS
where a is the area of the orifice in square inches, f is a coeffi
cient, Ti is the absolute initial temperature in degrees Fahren
heit at the absolute pressure pi in the " reservoir or highpressure
side" and p 2 is the absolute discharge pressure, both in pounds
per square inch. When the discharge from the orifice is directly
into the atmosphere, p2 is obviously the barometric pressure.
Westcott's and Weisbach's experiments show that the values
of f are about 0.925 for equation (217) and about 0.63 for equa
tion (218).
For small pressures it is often desirable to substitute manom
eters for pressure gages. One leg of a Utube manometer can be
connected to the highpressure side of the orifice and the other leg
to the lowpressure side. Valves or cocks are sometimes inserted
between the manometer and the pipe in which the pressure is to
be observed for the purpose of " dampening" oscillations. This
practice is not to be recommended as there is always the possi
bility that the pressure is being throttled.* A better method is
to use a Utube made with a restricted area at the bend between
the two legs. This will reduce oscillations and not affect the
accuracy of the observations.
Discharge from compressors and the air supply for gas engines
are frequently obtained by orifice methods.
When pi — p 2 is small compared with pi, the simple law of
discharge f of fluids can be used as follows:
* Report of Power Test Committee, Journal A. S.M.E., Nov., 191 2, page 1695.
f If the density is fairly constant,
144 pi v£ 144 p 2 Vq2
s + 2g s i "2g'
where Vi is the velocity in feet per second in the "approach" to the orifice, and v is
the velocity in the orifice itself. Since Vi should be very small compared with v ,
Vq2 144 (p x  pa)
2g s '
Vo = V /:
w
2 g X 144 (p x  p 2 ) .
= favosa fas y ^^fop.) ,
FLOW OF AIR THROUGH AN ORIFICE 157
fa
w = V 2 g X 144 (Pi  P2) s, (219)
144
where f is a coefficient from experiments, g is the acceleration
due to gravity (32.2), and s is the unit weight of the gas meas
ured, in pounds per cubic foot, for the average of the initial and
final conditions of temperature and pressure. If the difference
in pressure is measured in inches of water h with a manometer,
then
144 (pi — p 2 ) = — — X h (expressed in terms of pounds per square
foot),
w = y 2 ghs X — — (pounds per second), (220)
144 ^2
where 62.4 is the weight of a cubic foot of water (density) at
usual "room" temperatures.
This equation can also be transformed so that a table of the
weight of air is not needed, since by elementary thermodynamics
144 pv = 53.3 T, where v is the volume in cubic feet of one
pound and T is the absolute temperature in Fahrenheit. Since
v is the reciprocal of s, then
s = 144 p + 53.3 T,
and w = 0.209 fa V/^ ( 221 )
Here p and T should be the values obtained by averaging the
initial and final pressures and temperatures. Great care should
be exercised in obtaining correct temperatures. For accurate
work, corrections of s for humidity must be made.*
For measurements made with orifices with a wellrounded
entrance and a smooth bore so that there is practically no Con
or w = fa VsTg x 144 (pi — p 2 )s.
Professor A. H. Westcott has computed from accurate experiments that the
value of the coefficient f in these equations is approximately 0.60.
* Tables of the weight of air are given on page 181 and tables of humidity on
page 368 in Moyer's Power Plant Testing (2d Edition).
158 FLOW OF FLUIDS
traction of the jet the coefficient f in equations (217) and (218)
may be taken as 0.98. In the rounding portion of the entrance
to such a nozzle the largest diameter must be at least twice the
diameter of the smallest section. For circular orifices with
sharp corners Professor Dalby * stated that the coefficient for
his sharpedged orifices in a thin plate of various sizes from 1 inch
to 5 inches in diameter was in all cases approximately 0.60;
and these data agree very well with those published by Durley.j
When p 2 J pi = 0.99 the values obtained with this coefficient
are in error less than J per cent; and when p 2 f pi = 0.93 the
error is less than 2 per cent.
Receiver Method of Measuring Air. None of the preceding
methods are adaptable for measuring the volume of air at high
pressures as in the case of measuring the discharge in tests of
air compressors. Pumping air into a suitably strong receiver
is a method often used. The compressor is made to pump
against any desired pressure which is kept constant by a regu
lating valve on the discharge pipe:
Pi and P 2 = absolute initial and final pressures for the test,
pounds per square inch.
Ti and T 2 = mean absolute initial and final temperatures, de
grees Fahrenheit.
Wi and w 2 = initial and final weight of air in the receiver,
pounds.
V = volume of receiver, cubic feet.
PiV = wRTi, and P 2 V = w 2 RT 2 , 'where R is the constant
53.3 for air, then weight of air pumped
L (h _ h\
i.3 VT, Tj
w 2  wi = —  (— — )• (222)
In accurate laboratory tests the humidity of the air entering
the compressor should be measured in order to reduce this
* Engineering (London), Sept. 9, 19 10, page 380, and Ashcroft in Proc. Insti
tution of Civil Engineers, vol. 173, page 289.
f Transactions American Society of Mechanical Engineers, vol. 27 (1905),
page 193.
FLOW OF VAPORS
159
weight of air to the corresponding equivalent volume at atmos
pheric pressure and temperature.
The principal error in this method is due to difficulty in measur
ing the average temperature in the receiver. Whenever practi
cable the final pressure should be maintained in the receiver at
the end of the test until the final temperature is fairly constant.
Flow of Vapors. In Fig. 45 suppose the sections A and B
are so proportioned that the velocity of the substance passing
section A is the same as that at section B. Such a condition
might arise in a calorimeter or in the expansion of ammonia
through a throttling or expansion
valve, as in an ice machine. The
pressure at A will be Pi which is
greater than the pressure at B of P 2 .
Fig. 50 represents the entropy dia
gram for such a condition. As the
pressure falls from Pi to P 2 the maxi
mum heat available to produce veloc
ity through the nozzle is the area
acde. The value of the quality, rep
resented bv the symbol x for the
substance after leaving the nozzle FlG ' S °'~ Q ^%^ T n ° Ve ~
corresponds to that of point c and
the area acde is the excess kinetic energy represented by the
increased velocity. This excess kinetic energy is destroyed by
coming into contact with the more slowly moving particles at
B and with the sides of the vessel. The area acde is equal to
(h a + x a L a ) — (h c + x c L c ) and the relation of x a to x c is obvi
ously adiabatic. The area ohdbf equals area oheag (thus the
heat content at b is the same as at a). The location of b can be
found as follows:
I 6 ' Pi
a
Id
c b
/ *"
h
TEnt.Diagram
9
f
Xb
Ha ~T~ XaL*a »^&
(223)
The curve shown in Fig. 51 represents the discharge of a mix
ture of steam and water (x = 0.6 at 100 pounds per square inch
absolute pressure) into a vessel having the pressures shown. The
i6o
FLOW OF FLUIDS
points on this curve cannot be determined by entropy tables.
At ioo pounds per square inch pressure the total heat of the wet
steam is h + 0.6 L or 298.5 + 0.6 X 887.6 = 831. 1 B.t.u. per
pound.
2.00
1.75
1.50
o
a
o
o
£ 1.25
o
o 1
is
^ 1 1.00
ra 03
a .75
.50
.25
4/
to/
7
10 20 30 40 50 60 70 80 90 100
Discharge Pressureln. lbs. per sq. in.
Fig. 51. — Discharge of Steam Under Various Pressures.
At 60 pounds per square inch absolute pressure the total heat
may be found by the use of the entropy diagram shown in
Fig. 52. The entropy values are taken directly from steam
tables. The entropy for the initial point is, then,
ab = 0.4748 + 0.6 X 1. 1273 = 1.1512.
The distance
dc = ab — 0.4279 = 0.7233,
and x at 60 pounds is :
dc 0.7233
= = 0.595.
1. 2154 1. 2154
The total heat at 60 pounds pressure is
h + 0.595 L = 262 4 + 0.595 X 914.3 = 805.4.
FRICTION LOSS IN A NOZZLE
161
The velocity of flow is
V2 x 32.2 x 778 (831. 1 — 805.4) = 1135 feet per second.
The volume of one pound is
0.016 (1.0 — 0.595) + 7.166 X 0.595 = 4.27 cubic feet,
and the weight per cubic foot is
4.27
= 0.2342 pound.
1.1273
1.1512
1.2154
TEnt.Diagram
The weight discharged per square
inch per second is
1135 X 0.2342 _ ,
— — ^ = 1. 8 s pounds.
144 F
Velocity of Flow as Affected by
Radiation. Fig. 53 shows the radia ,..,.,.....,
tion losses. The condition at entrance tropy Diagram for Calculating
is represented at a and the area acde the Weight of Discharge of
represents the quantity of heat lost team *
by radiation. Area aefg represents the velocity change while
the point e represents, the condition of the moving substance.
If, after passing through the nozzle,
the velocity is reduced to that of en
trance, a point located as at b will
represent the condition of the sub
stance. This point would be so located
that
area aefg . , x
(224)
T. Ent. Diagram
eb =
L f
Friction Loss in a Nozzle. Fig. 54
shows the friction loss. The energy
converted into heat by friction varies
with the square of the velocity. In
Fig. 53. — Diagram illustrating this figure, a is the initial condition
Radiation Loss in Nozzle. and acfg ^ ^ energy available for
change in velocity provided there is no frictional loss. The ratio
of the areas acde to acfg is the proportional loss by friction.
The point c represents the condition of the substance at the
l62
FLOW OF FLUIDS
return of the friction heat to the substance. The heat is
returned in exactly the same way as if it came from an outside
source. The distance ch is the area
acde T L c . The area edfg represents
the energy expended in the velocity
change and the point h represents the
state of the substance at the point of
discharge.
This condition is found existing in
the fixed nozzle of most turbines.
Point a represents the condition on
the highpressure side of the nozzle
and point h, the lowpressure side.
The absolute velocity of discharge is
T. Ent. Diagram
n I d
a
Fig. 54. — Diagram Illustrating really caused by the energy repre
Friction Loss in Nozzle. S ented by the area edfg.
Impulse Nozzles. Suppose that the substance is discharged
with an absolute velocity corresponding to the area edfg (Fig.
54), and that it passes into a mov
ing nozzle, having the same pressure
on the discharge as on the intake
side. The energy represented by
the area edfg would be used up in
the following ways: (1) by the fric
tion in moving nozzle; (2) residual
absolute velocity; (3) and energy
used in driving the moving nozzle
against the resistance.
Fig. 55 shows the quantities used
up by each. Point a represents the
condition of the substance before
passing into the fixed nozzle while
point h shows its condition leaving
the fixed nozzle, the velocity corre
sponding to the area edfg. Area klde represents the energy
used up in friction in the moving nozzle; area klnm residual
T. Ent. Diagram
Fig. 55. — Diagram of Heat Losses
in a Steam Nozzle (Impulse).
TURBINE LOSSES
163
velocity after leaving moving nozzle and area mnfg represents
useful work used in moving the nozzle against its resistance.
The condition of the substance leaving the nozzle is shown at
q and not at h, the distance h r q being the area edlk divided
by Ln. The substance leaves the moving nozzle with a velocity
corresponding to the area klnm and it will have done work
corresponding to the area mnfg.
Turbine Losses. Fig. 56 is a simple velocity diagram show
ing, for an impulse nozzle such as occurs in many turbines, the
relative value of those various losses. A is a stationary nozzle
e v
Fig. 56. — Impulse Nozzle and Velocity Diagrams.
discharging against the movable blades B. The path of the steam
is shown by the dotted line. The line db marked v represents
the velocity of discharge of the stationary nozzle, which makes
an^angle a with the direction of motion of the moving blades.
Call ev the velocity of the moving blades, then h is the amount
and direction of the relative velocity of the steam over the
surface of the moving blades. It loses a portion of this velocity
as it passes over the surface of the blades and Ih becomes the
actual relative velocity of discharge. The direction of Ih is de
termined by the discharge edge of the moving blades, the angles
a and /3 being as shown. The residual absolute velocity is rep
resented by r.
The total energy equivalent of the velocity developed
in B.t.u. per pound =
2g
(225)
164
FLOW OF FLUIDS
w
The residual energy per pound = —
2 i
(226)
n
T. Ent. Diagram
Reaction Nozzles. When the substance leaving the station
ary nozzle passes into a moving nozzle having the pressure at
the intake greater than at the discharge, the conditions differ
from those just discussed. The velocity in this case is changed
in passing through the moving nozzle. In the equations given
in the previous discussion it was assumed that the moving nozzles
were entirely filled with the sub
stance, and when partly filled in the
expanding portion, coefficients of
correction were applied, but in this
case the nozzles should be so de
signed that the substance entirely
fills them, as the corrections are un
known.
In Fig. 57 the lines of Fig. 55 are
reproduced together with those re
lating directly to the reaction nozzle.
Point a corresponds to the condition
on entering the stationary nozzle,
Fig. 57. Diagram of Heat Losses •<. h ^ condition on leaving ft
in a Steam Nozzle (Reaction). .
with a velocity corresponding to the
area edfg. Point k represents the pressure at the discharge end
of the moving nozzle, and if no friction losses or impact loss occur
in the moving nozzle, point k would represent the condition of the
discharged substance and the area egmkhd would be accounted
for as useful work done and residual velocity. But since friction
losses and impact loss do occur a portion of this area edhkno can
be set aside to represent these losses, a portion noqp represents
the residual velocity, while the remaining area pqgm represents
the useful work done.
The condition of the substance leaving the moving nozzle is
, , ,, ,. . , , 1 • area edhkno .
given by k, the distance kl being Area onpq rep
Li
resents the residual velocity.
INJECTORS 165
Coefficient of Flow. Few experiments have been carried on for
determining the flow of steam in nozzles proportioned for maxi
mum discharge. For a nozzle having a well rounded entrance
and with the parallel portion of least diameter from 0.25 to 1.5
times the length of the converging entrance, the coefficient of
discharge is about 1.05. For properly shaped entrances and for
areas of orifices between 0.125 square inch and 0.75 square inch
the coefficient of discharge varies from 0.94, the two pressures
being nearly alike, to unity, the ratio of the pressures being 0.57.
For an orifice through a thin plate the coefficient is about 0.82,
the ratio of the pressures being 0.57.
Injectors. In an injector, steam enters at A in Fig. 58 at
the pressure of the supply. The quantity of water entering at
C, the crosssection of the pipe, and the pressure of the water
determine the pressure at B. At D
the pressure should be zero (atmos
pheric) or equal to the pressure in
the water supply pipe to which D Fig. 58. Essential Parts of an
may be connected. 'The total hy
draulic head should exist as velocity head at this point. At E
the pressure should be sufficient to raise the check valve into
the boiler and the velocity sufficient to carry the intended
supply into the discharge pipe.
The shape of the nozzle from A to B should be such as to con
vert the energy in the steam at A into velocity at B. At B the
water and steam meet, condensing the steam, heating the water
and giving to the water a velocity sufficient to carry it through
the nozzle BD.
All the energy accounted for at A and C must be accounted for
at E. The heat lost by radiation may be neglected. The veloc
ity at any section of the nozzle equals
volume passing in cubic feet
area of section in square feet
or V A = — » where a = area at any section corresponding to
velocity V and a a = area at section A .
1 66 FLOW OF FLUIDS
Weight of Feed Water Supplied by an Injector per Pound of
Steam. Assuming the steam supply to be dry the heat units
contained in the steam and feed water per pound and the heat
in the mixture of steam and feed water per pound may be easily
calculated.
Knowing the rise of temperature of the water passing through
the injector and neglecting radiation losses, the pounds of feed
water supplied per pound of steam used by the injector may be
obtained. Thus :
Heat units lost by steam = Kinetic energy of jet + Heat
units gained by feed water.
The kinetic energy of jet may be neglected since it is very
small, then,
H — h/ = w (km — hf)
H h f , v
W = => (227)
rim — hf
where w = the weight of feed water lifted per pound of steam.
h m = heat of liquid of mixture of condensed steam and
feed water.
hf = heat of liquid of entering feed water.
Thermal Efficiency of Injector. The thermal efficiency of an
injector neglecting radiation losses is unity. All the heat ex
pended is restored either as work done or in heat returned to the
boiler.
Mechanical Efficiency of Injector. The mechanical work per
formed by the injector consists in lifting the weight of feed water
and delivering it into the boiler against the internal pressure.
The efficiency, considering the injector as a pump, is
Work done
B.t.u. given up by steam to perform the work
or
where U = [wl s + (w + 1) l P ] f 778 (in heat units).
FLOW OF STEAM THROUGH NOZZLES
167
Is =
W =
pressure head corresponding to boiler gage pressure,
in feet,
suction head in feet,
pounds of water delivered per pound of steam.
Orifice Measurements of the flow of steam are sometimes used
for ascertaining the steam consumption of the "auxiliaries" in a
power plant. This method commends itself particularly because
of its simplicity and accuracy. It is best applied by inserting a
plate § inch thick with an orifice one inch in diameter, with square
edges, at its center, between the two halves of a pair of flanges
on the pipe through which the steam passes. Accurately cali
brated steam gages are required on each side of the orifice to
determine the loss of pressure. The weight of steam for the
various differences of pressure may be determined by arranging
the apparatus so that the steam passing through the orifice will
be discharged into a tank of water placed on a platform scales.
The flow through this orifice in pounds of dry saturated steam
per hour when the discharge pressure at the orifice is 100 pounds
by the gage is given by the following table:
Pressure drop,
lbs. per sq. in.
Flow of dry steam
per hour, lbs.
Pressure drop,
lbs. per sq. in.
Flow of dry steam
per hour, lbs.
1
2
I
2
3
4
430
6i5
930
1200
1400
5
10
15
20
1560
2180
2640
3050
Flow of Steam through Nozzles. The weight of steam dis
charged through any welldesigned nozzle with a rounded inlet,
similar to those illustrated in Figs. 59 and 60, depends only on
the initial absolute pressure (Pi), if the pressure against which
the nozzle discharges (P 2 ) does not exceed 0.58 of the initial
pressure. This important statement is well illustrated by the
following example. If steam at an initial pressure (Pi) of 100
pounds per square inch absolute is discharged from a nozzle, the
i68
FLOW OF FLUIDS
weight of steam flowing in a given time is practically the same
for all values of the pressure against which the steam is dis
charged (P 2 ), which are equal to or less than 58 pounds per square
inch absolute.
If, however, the final pressure is
more than 0.58 of the initial, the weight
of steam discharged will be less, nearly
in proportion as the difference between
the initial and final pressures is re
duced.
The most satisfactory and accurate
formula for the " constant flow " con
dition, meaning when the final pressure
is 0.58 of the initial pressure or less, is
the following, due to Grashof,* where w
is the flow of steam f (initially dry
saturated) in pounds per second, A Q is the area of the smallest
section of the nozzle in square inches, and Pi is the initial abso
lute pressure of the steam in pounds per square inch,
Fig. 59. — Example of a Well
designed Nozzle.
or, in terms of the area,
w =
A, =
ApPj
60
60 w
97
Pi
97
(229)
(230)
* Grashof, Theoretische Maschinenlehre, vol. 1, iii; Hiitte Taschenbuch, vol. 1,
page 2>2>Z Grashof states the formula,
w = 0.01654 A oPr 9696 ,
but the formula given in equation (229) is accurate enough for all practical uses.
t Napier's formula is very commonly used and is accurate enough for most
calculations. It is usually stated in the form
w =
70
where w, Pi, and Ao have the same significance as in Grashof s formula. The
following formula is given by Rateau, but is too complicated for convenient use:
w = 0.001 A0P1 [15.26 — 0.96 (log Pi + log 0.0703)].
Common or base 10 logarithms are to be used in this formula.
FLOW OF STEAM THROUGH NOZZLES
169
These formulas are for the flow of steam initially dry and
saturated. An illustration of their applications is given by the
following practical example.
Example. The area of the smallest section (A ) of a suitably
designed nozzle is 0.54 square inch. What is the weight of the
DeLaval Type.
Nozzle*
Diaphragm,
Curtis Type.
Fig. 60. — Examples of Standard Designs of Nozzles.
flow (w) of dry saturated steam per second from this nozzle when
the initial pressure (Pi) is 135 pounds per square inch absolute
and the discharge pressure (P 2 ) is 15 pounds per square inch ab
solute?
Here P 2 is less than 0.58 Pi and Grashof's formula is applicable,
170 FLOW OF FLUIDS
O.54 (l3$) 97
or, w = D ^ ) OJ/ >
60
0.54 X 1 16.5* , ,
w = __^z — ^l_ = L040 pounds per second.
60
When steam passes through a series of nozzles one after the
other as is the case in many types of turbines, the pressure
is reduced and the steam is condensed in each nozzle so that
it becomes more moist each time. In the lowpressure nozzles
of a turbine, therefore, the steam may be very wet although
initially it was dry. Turbines are also sometimes designed to
operate with steam which is initially wet, and this is usually the
case when lowpressure steam turbines are operated with the
exhaust from noncondensing reciprocating engines. In all these
cases the nozzle area must be corrected for the moisture in the
steam. For a given nozzle the weight discharged is greater
for wet than for dry steam; but the percentage increase in the
discharge is not nearly in proportion to the percentage of mois
ture as is often stated. The general equation for the theoretic
discharge (w) from a nozzle" is in the form f
* The flow (w) calculated by Napier's formula for this example is w = ^ —
70
= 1. 04 1 pounds per second.
f The general equation for the theoretic flow is
/ r 2 g+1 1
where the symbols w, A , Pi, and g are used as in equations (228) and (229). P 2
is the pressure at any section of the nozzle, vi is the volume of a pound of steam at
the pressure Pi, and k is a constant. The flow, w, has its maximum value when
2 ifl
I
mm
is a maximum. Differentiating and equating the first differential to zero gives
1
il = ( 2 V" 1
Pi U + i/ '
P 2 is now the pressure at the smallest section, and writing for .clearness P for
P 2 , and substituting this last equation in the formula for flow (w) above, we have
FLOW OF STEAM 1 71
fp
w = K K /—, (231)
where Pi is the initial absolute pressure and Vi is the specific vol
ume (cubic feet in a pound of steam at the pressure Pi). Now,
neglecting the volume of the water in wet steam, which is a usual
approximation, the volume of a pound of steam is proportional
to the quality (xi). For wet steam the equation above becomes
then
(232)
The equation shows, therefore, that the flow of wet steam is
inversely proportional to the square root of the quality (xi).
Grashof's equations can be stated then more generally as
w = ^r_ <*»>
. 60 W A / V
= _ p^ — (234)
These equations become the same as (229) and (230) for the
case where X\ = 1.
Flow of Steam when the Final Pressure is more than 0.58 of
the Initial Pressure. For this case the discharge depends upon
the final pressure as well as upon the initial. No satisfactory
formula can be given in simple terms, and the flow is most easily
calculated with the aid of the curve in Fig. 61, due to Rateau.
Wm?M
Now regardless of what the final pressure may be, the pressure (P ) at the smallest
section of a nozzle (Aq) is always nearly 0.58 Pi for dry saturated steam. Making
then in the last equation P = 0.58 Pi and putting for k Zeuner's value of 1.135
for dry saturated steam, we may write in general terms the form stated above,
V Vi
where K is another constant. See Zeuner's Theorie der Turbinen, page 268 (Ed. of
1899)
172
FLOW OF FLUIDS
This curve is used by determining first the ratio of the final to
p
the initial pressure — , and reading from the curve the corre
sponding coefficient showing the ratio of the required discharge
to that calculated for the given conditions by either of the equa
tions (229) or (233). The coefficient from the curve times the
1.0
1
1
— '
_ qT
,9
lC
I
.8
S§' 7
1 1 6
* *
ja <3 .5
?
+3 CO
§ £ A
^ 2
& a
<3 n
© r° .3
O PH *°
.2
i
:o
uo
.8 .7 «£
Batio of FfnaT to Initial PressureJ^L ,
Pi
.5
Fig. 61. — Coefficients of the Discharge of Steam when the Final Pressure is
Greater than 0.58 of the Initial Pressure.
flow calculated from equations (229) or (233) is the required re
sult. Obviously the discharge for this condition is always less
than the discharge when the final pressure is equal to or less
than 0.58 of the initial.
Length for Nozzles. The length of the nozzle is usually made
to depend only on the initial pressure. In other words, the
length of a nozzle for 150 pounds per square inch initial pressure
is usually made the same for a given type regardless of the final
pressure. And if it happens that there is crowding for space,
one or more of the nozzles may be made a little shorter than
the others.
UNDER AND OVEREXPANSION
m
Designers of De Laval nozzles follow practically the same
" elastic " method. The divergence of the walls of noncon
densing nozzles is about 3 degrees from the axis of the nozzle,
and condensing nozzles for high vacuums may have a divergence
of as much as 6 degrees * for the normal rated pressures of the
turbine.
One of the authors has used successfully the following empirical
formula to determine a suitable length, L, of the nozzle between
the throat and the mouth (in inches) :
L = V I5 A Q j
(235)
where A is the area at the throat in square inches.
Under and Overexpansion. The best efficiency of a nozzle is
obtained when the expansion required is that for which the nozzle
ao
$0
P « °
'» &
o *
« I
So
.2 «
1*2
*
—
r+
/
/
1
1
1
i
1
i
1
4
30 25 20 15 10 ^ 5
"Percentage Nozzle is too Small^
at Mouth (Under Expansion)
5 10 15 20 25 30
Percentage Nozzle is too Larger
at Month (Over Expansion)
Fig. 62. — Curve of Nozzle Velocity Loss.
was designed, or when the expansion ratio for the condition of
the steam corresponds with the ratio of the areas of the mouth
and throat of the nozzle. A little underexpansion is far better,
however, than the same amount of overexpansion, meaning
that a nozzle that is too small for the required expansion is more
* According to Dr. O. Recke, if the total divergence of a nozzle is more than
6 degrees, eddies will begin to form in the jet. There is no doubt that a too rapid
divergence produces a velocity loss.
f Moyer's Steam Turbines.
174
FLOW OF FLUIDS
efficient than one that is correspondingly too large.* Fig. 62
shows a curve representing average values of nozzle loss  used
to determine discharge velocities from nozzles under the condi
tions of under or overexpansion.
Nonexpanding Nozzles. All the nozzles of Rateau steam
turbines and usually also those of the lowpressure stages of Curtis
turbines are made nonexpanding; meaning, that they have the
same area at the throat as at the mouth. For such conditions
it has been suggested that instead of a series of separate nozzles
in a row a single long nozzle might be used of which the sides
.*■*! 1(U9
J6.19£H
Fig. 63. — Nonexpanding Nozzles.
were arcs of circles corresponding to the inside and outside pitch
diameters of the blades. Advantages would be secured both on
account of cheapness of construction and because a large amount
of friction against the sides of nozzles would be eliminated by
omitting a number of nozzle walls. Such a construction has not
proved desirable, because by this method no wellformed jets
are secured and the loss from eddies is excessive. The general
statement may be made that the throat of a welldesigned nozzle
should have a nearly symmetrical shape, as for example a circle,
a square, etc., rather than such shapes as ellipses and long rect
* It is a very good method to design nozzles so that at the rated capacity the
nozzles underexpand at least 10 per cent, and maybe 20 per cent. The loss for
these conditions is insignificant, and the nozzles can be run for a large overload
(with increased pressures) in nearly all types without immediately reducing the
efficiency very much.
f C. P. Steinmetz, Proc. Am. Soc. Mech. Engineers, May, 1908, page 628; J.
A. Moyer, Steam Turbines.
MATERIALS FOR NOZZLES 1 75
angles. The shape of the mouth is not important. In Curtis
turbines an approximately rectangular mouth is used because
the nozzles are placed close together (usually in a nozzle plate
like Fig. 60) in order to produce a continuous band of steam;
and, of course, by using a section that is rectangular rather than
circular or elliptical, a band of steam of more nearly uniform
velocity and density is secured.
Fig. 63 shows a number of designs of nonexpanding nozzles
used by Professor Rateau. The length of such nozzles beyond
the throat is practically negligible. Curtis nonexpanding noz
zles are usually made the same length as if expanding and the
length is determined by the throat area.
Materials for Nozzles. Nozzles for saturated or slightly super
heated steam are usually made of bronze. Gun metal, zinc
alloys, and delta metal are also frequently used. All these
metals have unusual resistance for erosion or corrosion from the
use of wet steam. Because of this property as well as for the
reason that they are easily worked with hand tools * they are
very suitable materials for the manufacture of steam turbine
nozzles. Superheated steam, however, rapidly erodes all these
alloys and also greatly reduces the tensile strength. For nozzles
to be used with highly superheated steam, cast iron is generally
used, and except that it corrodes so readily is a very satisfactory
material. Commercial copper (about 98 per cent) is said to
have been used with a fair degree of success with high super
heats; but for such conditions its tensile strength is very low.
Steel and cupronickel (8 Cu + 2 Ni) are also suitable materials,
and the latter has the advantage of being practically noncor
rodible.
The most important part of the design of a nozzle is the deter
mination of the areas of the various sections — especially the
smallest section, if the nozzle is of an expanding or diverging
type. In order to calculate the areas of nozzles we must know
how to determine the quantity of steam (flow) per unit of
time passing through a unit area. It is very essential that the
* Nozzles of irregular shapes are usually filed by hand to the exact size.
176 FLOW OF FLUIDS
nozzle is well rounded on the " entrance " side and that sharp
edges along the path of the steam are avoided. Otherwise it
is not important whether the shape of the section is circular,
elliptical, or rectangular with rounded corners.
Whether the nozzle section is throughout circular, square, or
rectangular (if these last sections have rounded corners) the
efficiency as measured by the velocity will be about 96 to 97 per
cent, corresponding to an equivalent energy efficiency of 92 to 94
per cent.
PROBLEMS
1. Air at a temperature of ioo° F. and pressure of 100 lbs. per sq. in. ab
solute flows through a nozzle against a back pressure of 20 lbs. per sq. in.
absolute. Assuming the initial velocity to be zero, what will be the ve
locity of discharge?
2. If the area at the mouth of the above nozzle is 0.0025 sq. ft. and the
coefficient of discharge is unity, how many pounds of air will be discharged
per minute?
3. What will be the theoretical kinetic energy per minute of the above
jet assuming no frictional losses?
4. Steam at a pressure of 150 lbs. per sq. in. absolute and 3 per cent
moisture flows through a nozzle against a back pressure of 17 lbs. per sq. in.
absolute. Calculate the velocity at the throat of the nozzle. What will
be the velocity at the end of the nozzle?
5. Steam at a pressure of 200 lbs. per sq. in. absolute, temperature
530 F., expands in a nozzle to a vacuum of 28.5 ins. (30in. barometer).
Calculate the absolute velocity of the steam leaving the nozzle.
6. The area of a nozzle at its smallest section is 0.75 sq. in. It is sup
plied with dry saturated steam at 160 lbs. per sq. in. absolute pressure and
the exhaust pressure is 2 lbs. per sq. in. absolute. How many pounds of
steam per hour will the nozzle discharge?
7. A steam nozzle is to be designed that will discharge 500 lbs. of steam
per hour. The pressure of the steam is 175 lbs. per sq. in. absolute and has
ioo° F. superheat. The exhaust pressure is 28 ins. vacuum (barometer
29.82). What will be the area of the nozzle at its smallest section and at its
end?
8. A safety valve is to be designed for a 200 horse power boiler generating
steam at 150 lbs. per sq. in. absolute pressure and 5 per cent moisture.
Assuming that the safety valve should have a capacity such that it will
release the boiler of all steam when generating double its rated capacity,
what will be the smallest sectional area of the valve?
PROBLEMS 177
9. An injector supplies water to a boiler. The boiler pressure is 100 lbs.
per sq. in. absolute and the steam generated is dry and saturated, the tem
perature of the entering feed water is 6o° F. and the temperature of the dis
charged mixture of condensed steam and feed water is 180 F. The suction
head is 3 ft. What is the weight of feed water supplied to the boiler per
pound of steam? What is the mechanical efficiency of the injector?
10. Design a nozzle to deliver 400 lbs. of steam per hour, initial pressure
175 lbs. per sq. in. absolute, final pressure atmospheric (barometer 28.62
ins.), temperature of steam 6oo° F.
CHAPTER X
APPLICATIONS OF THERMODYNAMICS TO COMPRESSED AIR AND
REFRIGERATING MACHINERY
COMPRESSED AIR
Air when compressed may be used as the working medium in
an engine, in exactly the same way as steam. Furthermore, it
is an agent for the transmission of power and can be distributed
very easily from a central station for the purpose of driving
engines, operating quarry drills and various other pneumatic
tools.
Air Compressors. The type of machine used for the compres
sion of air is that known as a pistoncompressor and consists of
a cylinder provided with valves and a piston.
The work performed in the aircylinder of a compressor can
best be studied successfully from an indicator diagram. If the
compression is performed very slowly in a conducting cylinder,
so that the air within may lose heat by conduction to the atmos
phere as fast as heat is generated by compression, the process
will in that case be isothermal, at the temperature of the atmos
phere. Also if the compressed air is distributed to be used in
compressed air motors * or engines without a change of temper
ature, and that the process of expansion in the compressed air
motors or engines is also indefinitely slow and consequently
isothermal, then (if we neglect the losses caused by friction in
pipes) there would be no waste of power in the whole process of
transmission. The indicator diagram would then be the same
per pound of air in the compressor as in the air motor or engine,
although the course of the cycle would be the reverse — that is,
it would retrace itself.
* Compressed air motors are similar to steam engines, but use compressed air
instead of steam.
178
WORK OF COMPRESSION
179
Adiabatic compression and expansion take place approximately
if the expansion and compression are performed very quickly,
or when the air is not cooled during compression, — in such
a case the temperature of the air would rise. The theoretical
indicator diagram of the compressor, Fig. 64, is FCBE and that
3Y
Volume
Fig. 64. — Diagram of Compressor.
Volume
Fig. 65. — Diagram of Air Engine.
of the air engine, Fig. 65, is EADF. CB and AD are both
adiabatic lines. The change of volume of the compressed air
from that of EB to EA occurs through its cooling in the dis
tributing pipes, from the temperature produced by adiabatic
compression down to the temperature of the atmosphere.
Suppose both diagrams of the compressor and of the air engine
are superimposed as in Fig. 66, and then an imaginary isothermal
line is drawn between the points A and C.
P
D
Volume
FlG. 66. — Superimposed Diagrams of Figs. 21 and 22.
It is then evident that the use of adiabatic compression causes
a waste of power which is measured by the area ABC, while the
use of the adiabatic expansion in the air engine involves a further
waste, shown by the area ACD.
Work of Compression. Assuming no clearance in the com
pressor and isothermal compression, the pressurevolume diagram
will be similar to Fig. 64. The work of the cycle will be
l8o COMPRESSED AIR AND REFRIGERATING MACHINERY
W = PcVc ~ PcVdoge^
V b
PbV
bV b,
which becomes, since P C V,
W =
P b V b ,
Vc
Vb
\ PcVAoge^.
V c
(236)
(237)
In practice the compression cannot be made strictly isother
mal, as the operation of the piston would be too slow. The dif
ference between isothermal and adiabatic compression (and ex
pansion) can be shown graphically as in Figs. 67 and 68. In
Volume
Fig. 67. — Compression Diagram.
Volume
Fig. 68. — Expansion Diagram.
these illustrations the terminal points are correctly placed for a
certain ratio for both compression and expansion. Note that in
the compressing diagram (Fig. 67) the area between the two
curves, ABC, represents the work lost in compressing due to heat
ing, and the area between the two curves, ACMNF (in Fig. 68),
shows the work lost by cooling during the expansion. The
isothermal curve AC will be the same for both cases.
The temperature of the air is prevented as far as possible
from rising during the compression by injecting water into the
compressing cylinder, and in this way the compression curve will
change. The curves which would have been PV = a constant, if
isothermal, and PV 14 = a constant, if adiabatic, will be very much
modified. In perfectly adiabatic conditions the exponent u n"
= 1 .40 for air, but in practice the compressor cylinders are water
jacketed, and thereby part of the heat of compression is con
ducted away, so that " n" becomes less than 1.40. This value
of " n" varies with conditions; generally the value is between
1.2 and 1.3.
EFFECT OF CLEARANCE UPON VOLUMETRIC EFFICIENCY 181
When the compression curve follows the law, PV n equals a
constant, work of compression (Fig. 64) is
w = p c v c + PcVc ~ PbVb  P b V b
n — 1
(PcVc  P b V b ), (238)
n — 1
1
since
P C V» = P»V»*; V b = V c (Q" (239)
Combining equations (238) and (239),
W^P.V.[i[^) J (240)
The Effect of Clearance upon Volumetric Efficiency. It is
impossible to construct a compressor without clearance, con
sequently the indicator diagram differs from the ideal case. At
the end of the discharge stroke, the clearance volume is rilled with
compressed air. When the piston moves on its outward stroke,
the clearance air expands and the suction valves of the compres
sor will be held shut until the piston has moved a sufficient dis
tance to permit the entrapped air to expand to atmospheric
pressure. When that point is reached any further movement of
the piston opens the suction valves and external air is drawn into
the cylinder. Thus the entire stroke of the compressor piston
is not effective in pumping air. The ratio of the volume of air
pumped to the volume swept by the piston, or piston displace
ment of the cylinder, is termed volumetric efficiency.
Fig. 69 illustrates an ideal compressor diagram with clearance.
The air entrapped in the clearance space equals
v 3 = CV S ,
where V s = volume swept or piston displacement of the cylinder.
C = percentage of clearance.
The air in Fig. 69 expands to F 4 at which point the inlet valves
open. The air drawn into the cylinder is represented by the
182 COMPRESSED AIR AND REFRIGERATING MACHINERY
difference in volume between Vi and F 4 . This air, as well as the
clearance air, is compressed to point 2, while the compressed air
is discharged from points 2 to 3. Knowing the per cent of clear
ex,
v 3 =cy s
L I 3
ft
a\
i\
1 \
1 ^
1
1
1
1
1
1
v%
\4
\w
^^— _ 1
i< 1 —
1
u
v*
1
1
\f
1
t
J
f*—
v s
1
Volume
Fig. 69. — Ideal Air Compressor with Clearance.
ance, the volume swept by the piston (V s ), and the initial and
final pressure, the volumetric efficiency (Vet) may be deter
mined from the following equations:
P% (CV s ) n = PiVi".
1
V* = (£)CV S ,
since V x = V s + CV S ,
V 1  V, = V s + CV S
®y
CV S
i\ 1
i + Cii 
p s v
V s .
TWO STAGE COMPRESSION
183
Therefore volumetric efficiency is
Vet. =
V s
= i+C
11
im
(241)
The volumetric efficiency also decreases with the altitude, as
the weight of a cubic foot of air decreases as the altitude increases.
Two Stage Compression. The problem of economy, obviously,'
becomes one of abstracting the heat generated in the air during
the process of compression. As previously mentioned, this is
partially accomplished by waterjacketing the cylinders, and also
by water injection. Nevertheless, owing to the short interval
within which the compression takes place, and the comparatively
small volume of air actually in contact with the cylinder walls,
very little cooling really occurs. The practical impossibility of
proper cooling to prevent waste of energy leads to the alternative
of discharging air from one cylinder after partial compression has
been effected, into a socalled intercooler, intended to absorb
the heat generated during the first compression, and then
compressing the air to the final pressure in another cylinder.
Volume
Fig. 70. — Indicator Diagram of Twostage Air Compressor.
This operation is termed " twostage " compression and when
repeated one or more times for high pressures the term " multi
stage " compression applies.
Referring to Fig. 70 and assuming the compression in a two
stage compressor to be adiabatic for each cylinder, the compres
184 COMPRESSED AIR AND REFRIGERATING MACHINERY
sion curve is represented by the broken line ABDE ; the compres
sion proceeds adiabatically in the first or lowpressure cylinder
to B; the air is then taken to a cooler and cooled under practi
cally constant pressure until its initial temperature is almost
reached, and its volume reduced from HB to HD; it is then
introduced to the second or highpressure cylinder and com
pressed adiabatically along the line DE to the final pressure
condition that was desired. It is seen that the compression
curve approaches the isothermal line FA.* The isothermal con
dition is obviously desired and, in consequence, airmachines
are built to approach that condition as nearly as possible.
Referring to Fig. 70, the work of each stage from equation (240)
will be
W (1st stage) =
W (2d stage)
n
n —
n
r p °4  ffi) " )•
The total work of compression is
W (total) = W (1st stage) + W (2d stage)
n
n
PaVa
1—^7
n—V
+ PaV c
Bn
©'
With perfect cooling P a V a = P d V a ; also P 6 = P d ; P c = Pe, then
n— 1 / n— 1\ ""
W (total)
n
n
PaVa
<rH(rj<
The work of compression as expressed by equation (242) be
comes a minimum when
n— 1 «— 1
© " + (8) "
(243)
is a maximum. Since the initial and final pressures P a and P c
are fixed, the pressure in the receiver, P b , can be found by differ
* The line FE represents further cooling.
REFRIGERATING MACHINES OR HEAT PUMPS
185
entiating equation (243) and equating this differential to zero, or
d
dPi
or
nl
nll
m'Hpyi
1 — n\ Pi
n
n
n
ln
ft
ln
= 0,
,Pc n
/ "— 1 1 — 2» \
* LlLl (Pc n P b n )
.. P & 2 = PoPc (244)
For compressing air to high pressures three and four stage com
pressors are used.
REFRIGERATING MACHINERY
Refrigerating Machines or Heat Pumps. By a refrigerating
machine or heat pump is meant a machine which will carry heat
from a cold to a hotter body.* This, as the second law of
thermodynamics asserts, cannot be done by a selfacting proc
ess, but it can be done by the expenditure of mechanical work.
Any heat engine will serve as a refrigerating machine if it be forced
to trace its indicator diagram backward, so that the area of the
Volume
Fig. 71. — Pressure volume Diagram of Carnot Cycle.
diagram represents work spent on, instead of done by, the
working substance. Heat is then taken in from the cold body
and heat is rejected to the hot body.
* This statement is not at variance with our knowledge that heat does not
flow of itself from a cold body to a hotter body.
1 86 COMPRESSED AIR AND REFRIGERATING MACHINERY
Take the Carnot cycle, using air as working substance (Fig.
71), and let the cycle be performed in the order dcba, so that the
area of the diagram is negative, and represents work spent upon
the working substance. In the stage dc, which is isothermal
expansion in contact with the cold body R (as in Fig. 7, page 41),
the air takes in a quantity of heat from R equal to wRT 2 log e r
[equation (46)], and in stage ba it gives out to the hot body H a
quantity of heat equal to wRT\ log e r. There is no transfer of
heat in stages cb and ad. Thus R, the cold body, is constantly
being drawn upon for heat and can therefore be maintained at a
temperature lower than its surroundings. In an actual refriger
ating machine operating with air, the cold body R consists of a
coil of pipe through which brine circulates while " working " air
is brought into contact with the outside of the pipe. The brine
is kept, by the action of the machine, at a temperature below
3 2 F. and is used in its turn to extract heat by conduction from
the water which is to be frozen to make ice. The " cooler " H,
which is the relatively hot body, is kept at as low a temperature
as possible by means of circulating water, which absorbs the heat
rejected to H by the " working " air.
The size of an air refrigerating machine is very large as com
pared with its performance. The use of a regenerator, as in
Stirling's engine (Fig. 8), may be resorted to in place of the
two adiabatic stages in the Carnot cycle, with the advantage of
making the machine much less bulky. Refrigerating machines
using air as working substance, with a regenerator, were intro
duced by Dr. A. C. Kirk and have been widely used.* The
working air is completely enclosed, which allows it to be
in a compressed state throughout, so that even its lowest
pressure is much above that of the atmosphere. This makes
a greater mass of air pass through the cycle in each revolution
of the machine, and hence increases the performance of a
machine of given size. In all air refrigerating machines the
* See Kirk, On the Mechanical Production of Cold, Proc. Inst, of C. E., vol.
XXXVII, 1874. Also lectures on Heat and its Mechanical Applications, in the
same proceedings for 1884.
SYSTEMS OF MECHANICAL REFRIGERATION 187
temperature range must be high to produce a given refrigerating
effect.
In another class of refrigerating machines the working sub
stance, instead of being air, consists of a liquid and its vapor,
and the action proceeds by alternate evaporation under a low
pressure and condensation under a relatively high pressure. A
liquid must be chosen which evaporates at the lower extreme of
temperature under a pressure which is not so low as to make
the bulk of the engine excessive. Ammonia, ether, sulphurous
acid, and other volatile liquids have been used. Ether machines
are inconveniently bulky and cannot be used to produce intense
cold, for the pressure of that vapor is only about 1.3 pounds per
square inch at 4 F., and to make it evaporate at any tempera
ture nearly as low as this would require the cylinder to be exces
sively large in proportion to the performance. This would not
only make the machine clumsy and costly, but would involve
much waste of power in mechanical friction. The tendency of
the air outside to leak into the machine is another practical ob
jection to the use of so low a pressure. With ammonia a dis
tinctly lower limit of temperature is practicable: the pressures
are rather high and the apparatus is compact. Carbonic acid
has been used as a refrigerant in small machines. The objection
to carbonic acid is that the pressures are very high as compared
with ammonia (see Appendix). The critical temperature of
carbonic acid is less than 90 F.
Unit of Refrigeration. The capacity of a refrigerating ma
chine is usually expressed in tons of refrigeration or ice melting
effect per twentyfour hours. As the latent heat of fusion of
ice is about 144 B.t.u., the heat units withdrawn per ton of re
frigerating effect per twentyfour hours is
2000 X 144 = 288,000 B.t.u. .
Systems of Mechanical Refrigeration. The standard systems
of mechanical refrigeration are:*
(A) The denseair system, socalled because the air which is
* Lucke's Engineering Thermodynamics, page 1148.
1 88 COMPRESSED AIR AND REFRIGERATING MACHINERY
the medium is never allowed to fall to atmospheric pressure, so
as to reduce the size of the cylinders and pipes through which a
given weight is circulating.
(B) The compression system, using ammonia, carbon dioxide
or sulphur dioxide, and socalled to distinguish it from the third
A
I
Circulating J
Water I
g^
A
S
Cool Com
pressed Air
f
Hot Com
pressed Air
A
,
^
^
Expt
Cyl
<? .
F
Compressor
Cylinder
E
Engine
Cylinder
nder
!
j
\
>
Cold Low
Pressure Air
Warmed Low
Pressure Air
/
" \
V
1 < B

•
1 '
Brine Tank
1 ~
" j*
[Circulating Brine]
Warm i Pipes \lcold
j jBrine
Brine
Fig. 72. — Dense Air System of Refrigeration.
system, because a compressor is used to raise the pressure of the
vapor and deliver it to the condenser after removing it from the
evaporator.
(C) The absorption system, using ammonia, and socalled be
THE AIR SYSTEM OF REFRIGERATION 189
cause a weak water solution removes vapor from the evaporator
by absorption, the richer aqua ammonia so formed being pumped
into a highpressure chamber called a generator in communica
tion with the condenser, where the ammonia is discharged from
the liquid solution to the condenser by heating the generator,
to which the solution is delivered by the pump.
No matter what system is used, circulating water is employed
to receive the heat, the temperature of which limits the highest
temperature allowable in the system and indirectly the highest
pressure.
The Air System of Refrigeration. The dense or closed air
system is illustrated in Fig. 72, in which air, previously freed of
moisture, is continuously circulated. The engine cylinder E
furnishes power * to drive the compressor cylinder F. This
cylinder delivers hotcompressed air into a cooler A where it
is cooled, and then passed on to the expansion cylinder G (tan
demconnected to both, the compressor F and to the engine
cylinder E), which in turn sends cold lowpressure air first through
the refrigerating coils in the brine tank
B and then back to the compressor cylin
der F; thus the air cycle is completed.
The courses of the circulating water and
also of the brine are shown by the dotted
lines.
The denseair cycle in a pressure volume Fig. 73. — Pressure vol
diagram is represented in Fig. 73, in which um e Diagram of Dense Air
BC is the delivered volume of hotcorn Cycle of Ref ^re
pressed air; CM is the volume of cooled air admitted in the ex
pansion cylinder; MB the reduction in volume due to the water
cooler; MN the expansion; NA the refrigeration or heating of
the air by the brme, and AB the compression. This operation is
but a reproduction of that previously described.
The work W c expended in the compression cylinder F is DABC
* Since the work done by the expansion of the coolcompressed air is less than
that necessary for the compressing of the air taken from the brine coils through
the same pressure conditions, a means must be employed to make up for the dif
ference, and for this purpose the engine cylinder is used.
190 COMPRESSED AIR AND REFRIGERATING MACHINERY
(Fig. 73) ; that done by the expansion cylinder G is W e = DCMN.
The shaded area MB AN represents the work which must be sup
plied by the engine E.
If w pounds of air are passing through the refrigerating machine
per minute, the heat withdrawn from the cold room or absorbed
by the brine along NA (Fig. 73) is
Qna = wC P (t a — h). (245)
The work expended in compressing w pounds of air, assuming
poly tropic compression, is, by equation (240),
If the compression is adiabatic,
W c =wP a Va^ L 
7  I
since =? = ( — ) v also AR = C v — C v = C v I J
If the expansion is complete in the expansion cylinder, the work
done in expansion is
We=^(t m 'h). (247)
The net work of the engine E required to produce refrigeration
becomes
W = Wc  We = ^ (h ~ t a  t m + *„). (248)
Since, from equation (245),
w = 77^ (ta — t n ), equation (248) becomes
l/*& ~~ ta — tm ~j~ tn\
\ t a t n /
7—7 \/NA[tb — t a — tm ~T t n \
W = j( : — ; ) * ( 2 49)
THE VAPOR COMPRESSION SYSTEM OF REFRIGERATION 1 91
The Vapor Compression System of Refrigeration. The com
pression system for ammonia or similar condensable vapors is
shown in Fig. 74. The figure illustrates the essential mem
bers of a complete compression refrigerating system. B repre
High Pres> '
Bure Liquid
R
Ammonia
Receiver
Circulating
Y Water
I
w
Condenser
<■
>
High Pres
Asure Vapor
Compressor
Cylinder
E
Engine
Cylinder
Throttling or
Expansion Valve
Low Pressure Liquid
Low Pres
sure Vapor
t= — Circulating Brine
J Pipes I
Warm f \ Cold
Brine I j Brine
Fig. 74. — Compression System of Refrigeration.
sents the directexpansion coil in which the working medium is
evaporated; F, the compressor or pump, for increasing the pres
sure of the gasified ammonia; E, the engine cylinder, — the source
of power; W, the condenser, for cooling and liquefying the gasi
fied ammonia; and V, a throttling valve, by which the flow of
192 COMPRESSED AIR AND REFRIGERATING MACHINERY
liquefied ammonia under the condenser pressure is controlled as
it flows from the receiver R to the expansion coils; B, the brine
tank, in which a materially lower pressure is maintained by the
pump or compressor in order that the working medium may boil
at a sufficiently low temperature to take heat from and con
sequently refrigerate the brine which is already cooled.
The operation of the compressor F (Fig. 74) is theoretically a
reversed Rankine cycle (Figs. 31 and 32). If one pound of vapor
is passed through the system, the work (Wc) of the compressor is
W c = (#2  #1) (B.t.u.),
where
Hi = the total heat of the vapor at entrance to the com
pressor,
H 2 = the total heat of the vapor discharged from the com
pressor.
The vapor entering the compressor may be treated as though
it were dry and saturated. The entropytemperature diagram
(Fig. 75) illustrates this case. The vapor discharged by the
compressor will then be superheated, having a temperature
t s at a pressure P 2 . The condition of the discharged vapor is
determined by equating the entropies at the inlet and discharge
pressures. The total heat (H 2 ) of the discharged vapor will be
H 2 = [fa + L 2 + C P (t s  k)]. (250)
The horse power of the compressor is, if w pounds of vapor is
circulated per minute:
_ ^#0 778, ( j
33,000
The heat absorbed per minute by w pounds of vapor passing
through the refrigerating or brine coils will be
Q r = w (fa + Li  fa)
= w (Hi  fa). (252)
fa = the liquid heat of the vapor after leaving
the condenser.
THE VAPOR ABSORPTION SYSTEM OF REFRIGERATION 193
The heat absorbed per minute by the condenser (Q c ) is
w (H 2  ho) (B.t.u.). (253)
The heat absorbed by the condenser is theoretically equal to
the sum of the heat absorbed in the refrigerator and the heat
equivalent of the work of compression.
Ammonia compressors are operated either on the dry or on
the wet system. In the dry system the ammonia entering the
compressor is a dry vapor as illustrated in Fig. 75. In the wet
system the ammonia enters the compressor in the wet state, the
heat developed during the com
pression being used in evapo
rating the liquid ammonia into
a vapor.
The Vapor Absorption Sys
tem of Refrigeration. The ab
sorption system depends upon
the fact that anhydrous am
monia possesses the property
of forming aqua ammonia.
The amount of ammonia water
will absorb depends upon the
temperature of the water; the
colder the water, the greater are
its absorptive powers.
The absorption system differs
from the vapor compression
system in that the absorption system replaces the compressor
by an absorber, where the anhydrous ammonia is changed into
aqua ammonia, a pump which transfers the ammonia from the
absorber to the generator, and a generator, where the aqua
ammonia is heated. Both systems have a condenser where the
ammonia is cooled and liquefied and an expansion valve or
throttling valve by means of which the flow of liquid ammonia
to the expansion coils is controlled. In the absorption system
the anhydrous ammonia vapor flows from the expansion coils to
T 2
T/'
/ p 2 y
/ , \
Si
S
a
£
H
Fig. 75.
Entropy 0
— Entropy Temperature Dia
gram for Ammonia.
194 COMPRESSED AIR AND REFRIGERATING MACHINERY
the absorber, in which the anhydrous vapor comes in contact
with weak aqua ammonia. The weak ammonia absorbs the
ammonia vapor. From the absorber the ammonia is pumped
into the generator, where it is heated by steam coils. The vapor
driven off in the generator passes to the condenser and from there
through the expansion or throttle valve to the expansion coils
which are located in the brine tank or in the refrigerating room.
The absorption system is usually provided with a rectifier to
thoroughly dry the gas before it enters the condenser, an ex
changer which heats the strong ammonia by cooling the weak
ammonia, and with other auxiliary equipment to reduce the heat
losses in the system.
COEFFICIENT OF PERFORMANCE OF REFRIGERATING MACHINES
The Coefficient of Per 1 ___ Heat extracted from the cold body
formance j" Work expended
This ratio may be taken as a coefficient of performance in esti
mating the merits of a refrigerating machine. When the limits
of temperature 2\ and T 2 are assigned it is very easy to show by
a slight variation of the argument used in Chapter IV that no
refrigerating machine can have a higher coefficient of performance
than one. which is reversible according to the Carnot method.
For let a refrigerating machine S be driven by another R which
is reversible and is used as a heatengine in driving S. Then if S
had a higher coefficient of performance than R it would take from
the cold body more heat than R (working reversed) rejects to
the cold body, and hence the double machine, although purely
selfacting, would go on extracting heat from the cold body in
violation of the Second Law of Thermodynamics. Reversi
bility, then, is the test of perfection in a refrigerating machine
just as it is in a heatengine.
When a reversible refrigerating machine takes in all its heat,
namely Q c at T 2j and rejects all, namely Q a at T\, and if we repre
sent the heat equivalent of the work done by W = Q a — Qc, then
the coefficient of performance is, as already denned,
PROBLEMS 195
Qc _ Qc T 2
W QaQc ZVT,
(254)
Hence, the smaller the range of temperature, the better is the
performance. To cool a large mass of any substance through a
few degrees will require much less expenditure of energy than to
cool onefifth of the mass through five times as many degrees,
although the amount of heat extracted is the same in both
cases. If we wish to cool a large quantity of a substance it is
better to do this by the direct action of a refrigerating machine
working through the desired range of temperature, than to cool
a portion through a wider range and then let this mix with the
rest. This is only another instance of a wide, general principle,
that any mixture of substances at different temperatures is ther
modynamically wasteful because the interchange of heat between
them is irreversible. An icemaking machine, for example,
should have its lower limit of temperature only so much lower
than 3 2 F. as will allow heat to be conducted to the working
fluid with sufficient rapidity from the water that is to be frozen.
PROBLEMS
1. If 200 cu. ft. of free air per minute (sealevel) is compressed isother
mally and then delivered into a receiver, the internal pressure of which is
102.9 lbs. per sq. in. absolute, find the theoretical horse power required.
2. What will be the net work done in footpounds per stroke by an air
compressor displacing 3 cu. ft. per stroke and compressing air from atmos
pheric pressure to a gage pressure of 75 lbs.? (Isothermal compression.)
3. What horse power will be needed to compress adiabatically 1500
cu. ft. of free air per minute to a gage pressure of 58.8 lbs., when n equals
4. A compressedair motor without clearance takes air at a condition
of 200 lbs. per sq. in. (gage) and operates under a cutoff at onefourth
stroke. What is the work in footpounds that can be obtained per cubic
foot of compressed air, assuming free air pressure of 14.5 lbs. and n equal
to 1 .41?
5. Find the theoretical horse power developed by 3 cu. ft. of air per
minute having a pressure of 200 lbs. per sq. in. absolute, if it is admitted
and expanded in an air engine with onefourth cutoff. The value of n is
1.2. (Neglect clearance.)
196 COMPRESSED AIR AND REFRIGERATING MACHINERY
6. Compute the net saving in energy that is effected by compressing
isothermally instead of adiabatically 50 cu. ft. of free air to a pressure of
200 lbs. per sq. in. gage. Barometer = 14 lbs. per sq. in., and a tempera
ture of 70 F. What is the increase in intrinsic energy during each kind
of compression? How much heat is lost to the jacketwater during each
kind of compression?
7. Let a volume of 12 cu. ft. of free air be adiabatically compressed in a
one stage air compressor from atmospheric pressure (15 lbs.) to 85 lbs. gage;
the initial temperature of the air being 70 F.
(a) What is the volume and temperature of the air after the com
pression?
(b) Suppose this heated and compressed air is cooled to an initial
temperature of 6o° F. at constant volume, what is its pressure
for that condition?
(c) Now if the air occupies such a volume as found in (a) and at
an absolute pressure as determined in (b), at a temperature
of 6o° F., and is then allowed to expand adiabatically down
to atmospheric pressure (15 lbs.), what is the temperature
and volume of the expanded air in Fahrenheit degrees?
8. Compare the work required to compress one pound of air from atmos
pheric pressure and a temperature of 6o° F. to a pressure of 200 lbs. per sq.
in. absolute in (1) a onestage and (2) in a twostage compressor. Assume
n = 1.25.
9. Using equation 249 as a basis, deduce a formula for the horse power
required to abstract a given number of heat units per minute by an air
refrigerating machine.
10. Calculate the approximate dimensions of a tenton air refrigerating
machine and the power required to drive it. Pressure in the cold chamber
is atmospheric and the temperature 32 F. Pressure of air delivered by
compressor is 90 lbs. per sq. in. gage. The temperature of the air coming
from condenser is 85 F. and the machine operates at 60 r.p.m. Allow
12 lbs. drop in pressure between the compressor and the expanding cylinder.
11. Calculate the approximate dimensions of a tenton ammonia com
pression refrigerating machine and the power required to drive it. The
temperature in the expansion coils is 15 F. and the temperature in the con
denser is 8 5 F. The machine is double acting and operates at a speed of
80 r.p.m.
12. What is the icemaking capacity per twentyfour hours of the machine
in problem 11? The temperature of the water to be frozen is 8o° F.
APPENDIX
Table I
SPECIFIC HEAT OF GASES AND VAPORS
(Taken from Smithsonian Physical Tables)
Substance
Air.
Alcohol,
(ethyl)
Alcohol,
(methyl)
Ammonia .
Benzene.
Carbon
Dioxide
Carbon
Monoxide
Ether.
Hydrogen. .
Nitrogen
Sulphur
Dioxide
Water
Range of
Temp.° C.
3010
0100
0200
20100
Mean
108220
101223
23100
27200
24216
Mean
34iiS
35i8o
116218
2877
15100
11214
Mean
2399
26198
69224
27189
251 1 1
Mean
289
12198
21 — 100
Mean
0200
16202
128217
100125
Mean
Cv
0.23771
0.23741
0.23751
0.2389
0.23788
04534
0.4580
0.5202
o.5356
0.5125
0.5228
0.2990
0.3325
03754
0.1843
0.2025
0.2169
0.2012
0.2425
0.2426
•4797
.4618
.4280
■4565
•3996
.4090
.4100
.4062
.2438
■1544
.4805
3787
4296
Authority
Regnault
1 1
1 1
Wiedemann
Regnault
Regnault
Wiedemann
Regnault
Wiedemann
< <
Regnault
Regnault
Wiedemann
Regnault
Wiedemann
Regnault
Wiedemann
Regnault
Regnault
Regnault
Gray,
Macfarlane,
Cv
C V
1 .4066
i.i3 6
131
1.300
1403
1 .029
1 .410
1 .410
1 .26
1.300
Authority
Various
Jaeger, j
Neyreneuf /
Cazin,
Wiillner
Wiillner
Rontgen,
Cazin,
Wiillner
Muller
Cazin
Cazin
Cazin
Muller
Various
Calcu
lated
Co
O.1691
O.3991
O.3991
O.1548
O.1729
O.4436
2.419
0.1729
O.1225
0.3305
197
198
APPENDIX
Table II
DENSITY OF GASES
(Taken from Smithsonian Physical Tables)
Gas
Specific
gravity
Grams per
cubic centimeter
Pounds per
cubic foot
Air ,
Ammonia
I .OOO
0.597
I.529
O.967
O.320O.740
O.0696
1 .191
0.559
O.972
1 .105
2.247
O.OOI293
O.OOO770
O.OOI974
O.OO1234
. 0004140 . 00095 7
0.000090
0.001476
0.000727
0.001257
0.001430
0.002785
O.08071
O.04807
O.12323
O.07704
O.02583O.05973
0.00562
O.09214
0.04538
O.07847
O.08927
O.17386
Carbon dioxide. ......
Carbon monoxide
Coal gas
Hydrogen
Hydrogen sulphide. . .
Marsh gas
Nitrogen
Oxygen
Sulphur dioxide
APPENDIX
199
Density
Weight per
Cubic Foot,
Pounds.
3
OO H OMOO PO W On
rj voO lOV)iO^ W)H O O von rrO n 10 w On O
POO On cm 10OO m ^tf N O N^tOvO hi O m\0 O "^O
OOOhhihicmcmcmpo 1000 m POO 00 m POO 00 O
OOOOOOOOOOOOHIMMHICMCMCMCMPO
000000000000000000000
000000000000000000000
a!
p.
•»
PO Ovo NO 00 O O
OOOOOOOOOnOvovoio pOOO 10 cm co co m po
lO^H tj CM HI N CM N PO POOO O PO HI fONNOO lO CM
tO pi 't Q> ^ ^O HO f)N h OnNNiO'^'^pOpOpO
Omo O no ioi ^t^toH w
CM HI H
Si
+
OO NIOO CM Tj fO <M PO^OOO TJ CM lO HI POOO rj r t^~
N « NtO POOO 10 ^ ^ lOOO ^ M PO00 O IOICNOM
t^ M MOtOH O00 NHOOO ^j CM M 000 N N
mhiOOOOOOnOnOn Onoo 0000000000 j>r^.r^r^
CMCMCMCMCMCMCMHIHIHIHIHIHIHIH1HHHIHIHIH1
Entropy
of the
Vapor.
k]I^
O r} *0 O PO 10O NOO N h OO •^■■^•w O CM N 10O
VO O POPOH lorCtNN CO Tj M 00 M0000 O rr OO
N.H n ^ h aNin^ ^too rjOoo 10 po cm 000 n
w O O On On 000 00 00 00 r^O OO 10 10 10 10 10 "t ^
CMCMCMHIHIHIHIHIHIHIHIHIH1HIHHIHIHIHHIH
Entropy
of the
Liquid.
<£>
CM POO OO O^N !OlON O00 00 00 M O POO CM CM N
O CM TJ O CM CM HI OnO CM Tj O Oi<tNNM^f)00
O rJO OOOnOhihicmponOhipovo *0O NOO O On
OOOOOHHIHIHIIIHlCMCMCMCMCMCMCMCMCMCM
OOOOOOOOOOOOOOOOOOOOO
Heat
Equivalent
of
External
Work.
•^ 10O to co On rr O +N POOO O O 00 tt O O w tnoo
^O NOO OnOnOmwhtjvo NOO 00 On O O H H H
loioirjioio too OOOOOOOOO t^r^t^t^f>»
Heat
Equivalent
of"
Internal
Work.
a
POM N ^VO On NOO N ©N'tO "3 O ^ CM "tf O00 00
N lO N CM NfOO N to CM O O 00 CM n cm 00 tt O n ^
M Q On OnOO O000 NNNio^JrOrOM cm m m m O O
O On On On On On On On On On On O On On On On On On On On
M HI
+
II
&5
NOO OnO O M POCM O^OO IOIONIOO M H ON to
•sj CM N M <t N 0\H CM ^ioh<3 O POO On H PO ^O
NOO 00 OOnOnOnO O O w CM CM fOtOCOrO^  ^'<t'^ 
OOOOOOOmmhhmmhhmmmhmh
mmwmmhhhmmhhhmhmhhmmh
Heat of
Vaporiza
tion.
►J
NO POO O 00 w NO O O co I s * POOO OO CM O O cm O
M M IOO N CO HI 00 O Tf HI CM IOO IOMOO tr> CM OnO
f~0 lOlOTjT}^POPOPOCM m O O On OnOO OOOO NN
OOOOOOOOOOOOOOOnOnOnOnOnOOn
HHMHHHMMHHMhHMM
Heat of
the
Liquid.
•«
lOtOMOH TTOO tO H
O CM to On N CO hi Tf POOO O 'ct  On h On NOO N h N a
C)h N O N POOO HO On^OO O N <3 O O h to On
CM PO ^r ^" VO IOO O O "On O CM PO CO *3" 10 IOO
MMMMMHMMHM
Tempera
ture,
Degrees F.

CO to On h 00 cmoOO co co to cm mOOO too N cm too
O m tt OnO CO h <sf POOO hioOmOOOOOnmno
10 PO •"*■ CM On >0 O "300 MO m pocm OO CMOO ro N H
PO IOO N t>.00 On On On O CM tJ 10O !>• r^OO 00 On On O
Absolute
Pressure,
Pounds per
Square
Inch.
■fc.
h cm po ^f IOO N00 On
OOOOOOOOOmcmpOtJ 10O N00 0> O h cm
H — ■!■ ■!■ H M M
a
200
APPENDIX
<
ft
co.
P
ft
H
<
ft
ft
O
CO
ft
i— i
ft
ft
ft
O
ft
ft
3
<
Density
Weight per
Cubic Foot,
Pounds.
§
O 0\«\0 o
POO toOW rrOO m ^ ifl io to »t (O N m O^t^'sJHOO
PO to t^OO O^h N t toO f^OO a O m N N POrjiovo
PO PO PO fO ^O t^OO OO H N W) toO t»00 Os O m M
OOOOOOOOOmiiihmmmmmmw<n<n
ooooooooooooooooooooo
3 5
ft O
■a
Os
PO <N Q\ t»00 O Tj Os OS Os H 00 MOO w t>»0 Os to N
O O **» <N O CO t^OO i W5 10NHO MOO 't HOO^O ^
OOOOO O O pom O OsOO *•*• t^O O to to to tJ rj tJ>
PONNNNMHHM
el
+
"t <t ID On O^O H00 M lOH to <N 00 t> N O w to M O
O O O ■* <N <0 OsO O O 00 O COM3 O to O toOO <N
O O VO to PO HH OsOO NO lO to Tf PO PO <N M M M O O
r>» r» N N r» t>.vo OOOOOOOOOOOOOO
MHMHMMHMHHtHHHMMMHHWItM
Entropy
of the
Vapor.
•^Ih
Os po NO to ^t h O m •^oo ON O TfO 00 to M M N N
PO N >t h vO O hO rh rjO O O PO Os NO O O O N
so to tt tt OsO poOCOso ^pOih OOO NO to Tf PO N
^^r^TrPOCOCOCONNCSNNNMMHHHMH
Entropy
of the
Liquid.
<*>
lO HI OO f)>0« OOO O M POO N TfM TJ lO O ^" ^ PO
N 00 h PO to PO00 so <N N m Os n tj m NcoO't Oi^
O O M H PO tOSO OO^OHHNtTt^lO tOSO SO N
OOOOOOOOOOOOOOOOOOOOO
Heat
Equivalent
of
External
Work.
a
NO Os ON PO <0 <N C\^0 M SO OS NOO H lo N O N ION
« N <N csi rj tosO so NOO OOOOOsOsOOOmmmm
t>»tv.r^.t^t^t^r^t>»t>t^t^.i>.r^ noo oo oo oo oo oo oo
Heat
Equivalent
of
Internal
Work.
a
O <0O 000000 O w Os CO O NO rtPO tOOO PO O NO
<N Os t^.so tosO ON toO toO to m N co OsO PO OsO
O Os OS OS 00 i^soso iOto^^POPO<N N M M M O O
OsOO oooooooooooooooooooooooooooooooooooooo
's'i
+
II
O *3* Tf N <N ^ onoO ^Jso so rr O tooO 00 PO ^ tt ^ CO
00 Os O O O O tOO Os m PO to t^OO Os hh csi po Tf toO
rt rj to to toO OOO t^t^t^t>»r^ l>00 OO 00 00 00 00
WMHHHWHMMMHHMHMHIIHIIHIMH
HHHMHWMHH.HMMHHMMMHMMH
Heat of
Vaporiza
tion.
hj
M ON'Jtt^O O m OsPONtoO O^O N N(Oh OsOsO
Tf H O Os O <N IO00 POOO POOsrlM NtOO NtOOM
t^ t^ t^O OtorjPOPO<NCSHMMOOOOsOs OsOO
OsOsOsOsOsOsOsOsOsOsOsOsOsOsOsOs O^OO 00 00 00
Heat of
the
Liquid.
i«
00 to O O h ^00 Os h rj h PO h t^O ^ O PO to to PO
PO t^ O m O 00 00 i>0 POOO <N ion t^NO O TfOO
t^. l>00 OOOsOMNPO^to too O t*» t>O0 OO Os Os Os
HMHHMNNNNNNNNCNNNCSNNNCS
Tempera
ture,
Degrees F.

MOO
OOtoO O O m POPOPOtoO h t^ O OsO ^0 pO O 00
lO Os N POOO O O ON't H N N00 N t>NO O ^ t"*
O O h h n ^io toO t^oo OOOsOsOOmmnnn
NNNNNNN<NCSC>lNN<NCSPOPO<OCOfOfOCO
'
Absolute
Pressure,
Pounds per
Square
Inch
«.
pO^rftoO toO toO toO toO toO toO too too
MMMMCSNPOPO'^'^tO tOO O t^ t^OO 00 Os Os O
H
L
APPENDIX 20I
to cn t^crjw r^. cn n fiO to o cooo cooo cn t^» cn
CO t^ r^OO OOO O m <N CN CN to rO i" "t 10 lOO t*» r"» t^OO 00 On On O M rO 10
\0 rf too r^OO O m <N co •>^ i^O t^OO O O m cn co ■>*• 100 t^OO O cn ■<*■ O tj
<n CN cn cn (N cn cocOPOcococococo<Ocos}T}'!tTt'3<3<3'3TiTjtoto toO
o o o o o o
OOOOOOOOOOOOOOOOOOOOOOOO
o *■*» o o <o cn
CO ^loo cn co to
CN O oo j> to ■<*■
m On cn cn O rf co to cn CO00 O O O t^. r^OO H \0 TJ rj O O M
com m m CN toiONO COO O Tt" On COOO co On tJ O CN tOOO to
co cn m O QnCO r^O Oio^^cocncnmihOOO OnOO O to
Tf ^ tO to to to
COCOCOCOCNCN*CNCNCNCNCNCNCNCNCNCNCNCNCNCNMMMM
O cn r^ co On r*
00 ^O Nt^O
O O OnOO 00 00
to to to to to to
r^f^O\<N ^CNtoO I^coOOOOOOOOO Onm m OnO On On
l^ tJ m OnO co m OnO ^<N ONNiorOH On rO "t O r~~ On CN
t~~ r^ t^O OOO totototorJ<^Tirl^cococo<Oco<N M M
totototototototototototototototototototototototo
m 00 o ^ o o
On O co tooo O
w m O OnOO 00
M M M O O O
CN toCN O OnmnO hGO to'lrtO OncOOC to m 0~0CO Q Onm
•^ r^. m tooo cor^CNNO hO m no m t^CNOO ^ONtor^O m to
NOO io^^coco<N <N M M O O ON O00 00 t^t^OO "3" CN
OOOOOOOOOOOOOOOnOnOnOnOnOnOnOnOnOn
H 11 M M M M
mmmmmmmmmmmmmmOOOOOOOOOO
On 'sj r~» On O00 ion no tooo o\ on OnOO O t O N cooo tOOO cn O coO O 00
CO tON h lO 6fON O Tjt^O COVO On <N to00 hi COO O0 M COO 00 CO f^OO *>■
f>00 0O O Os ON O O M M M CM <N (N <N tOf)tO^  ^' , ti" 1 1/ 5 1 to O O t*» 00
^rf^rt^rtlOlOlOtOtotOtotOtotOtototolOtototOtotototovoiOlO
o o o o o o
OOOOOOOOOOOOOOOOOOOOOOOO
On h co too 00
On <N CO toO t^OO On O H CN CO "^ to toO t>» t^OO O HI COO
m CN cm CN cn cn cn cocOfOfOCOCTO(Oit , ti  i'i' , t't , t , J , t^' l/ ' l O , O l O
0OOOOOOO0O0OO00OO0COO0OOO00OOO00O00OO0O00O0OO0O0O0O0OO000O0O
»0 N aMO O
toO N^HOOO iO'^'!i , ^^^J too 00 O CO tOOO "cj CN 00
COO N lO N O MO N 000 to CO M O^N") fO H O^t^lO^CN O 00 tO<N Tj O
O O On O On OnOO 00 00 00 **» f> f*~ r^NO OOOO totototototortTjTfcoCN
cn 0000 cOOO cn 00 'dO toO rj On rtoO f^NH tooo cm O On cn On tooo h
t^00 00 On O m
00 00 OO 00 On On
m cn cn co^Tltoto too O r*» t^00 0000 O^ ON On O O m cn^
OnO^OnOnO^OnOnOnOnOnO^OnOnOnCnOnO^O^O^O o o o o
CN tOOO N Nt<N
OnO "<d <n O 00 00 t^ r>00 OO O O cn tJO On cn •rjoo io ^OO co
to cn ONt^^J"<N ONiorOHOOO tJ cn OOOO to co m On t^O ^ cn OnO 00 m
OOOO NNls r^O OOOO tototototo^TjTj'^ , ^co<Oco<OcOCN) cn ii m
ooooooooooooooooooooooooooooooooooooocoooooooocoooooocoooooo
O to,0 to too
f^O ^" CN ONO CN t^.CNNO O iNONH CN •^■rJtOlO'chCN CN t^
CN to Cn <n tooo
O O O M W M
CO CO CO CO CO co
H ^NO « tOOO O CO tOOO O <N TfN O^H fO>ONH lOTfCN
CN CN CN cOcO<OcO'^'^^'^totototo toO O O O t*~ t^OO On
cocococo<Ococococococococo<OCOcocococococotOcoco
tJ00 m CO ■* "*■
CO H 00 to O O O tOOO M •stO 00 On O O On OnOO \tH V)ifl
M tJOO h ^NO "5 tooo M COO OO O co to r^ On M ^O OO ONH !ON h ON t^
cococo^'^'^tototo too o o o i^ r^ t^ r~ r^co oooooooo a ONO>0 O m
cococococococococococococococococo<OCOcococ r 5COcocococO , 5)"^t  ' , 4'
to O to O to
O M M CN CN CO
toO <fl ioO too toO toO toO toO toO >oO O O toO
CO Tf ^ to toO O f^ t~^O0 OOOiO>OOhhN(NI«5tj>ONO
202
APPENDIX
Table IV. — PROPERTIES OF SUPERHEATED STEAM
Reproduced by permission from Marks and Davis' "Steam Tables and Diagrams."
(Copyright, 1909, by Longmans, Green & Co.)
Pressure,
Satu
Degrees of Superheat.
Pressure,
Pounds
rated
Pounds
Absolute.
Steam.
50
100
150
200
250
300
Absolute.
t
162.3
212.3
262 .3
312.3
362.3
412.3
462.3
/
5
V
733
797
857
91.8
978
103.8
109.8
V
5
h
"30 5
H535
1176.4
H995
1222 .5
1245.6
1268.7
h
t
193.2
243.2
293.2
3432
3932
4432
4932
t
10
V
38.4
4i5
44.6
477
50.7
537
56.7
V
10
h
1143.1
1166.3
1189.5
1212 .7
1236.0
12593
1282.5
h
t
213.0
263 .0
3I30
3630
4i3o
463.0
5i3
t
*5
V
26.27
28.40
30.46
32.50
3453
36.56
38.58
V
15
h
1150.7
1174.2
1197.6
1221 .0
1244.4
1267 .7
1291 .1
h
t
228.0
278.0
328.O
378.0
428.0
478.0
528.0
t
20
V
20.08
21 .69
2325
24.80
26.33
2785
2937
V
20
h
1156.2
1179.9
1203.5
1227. 1
1250.6
12741
1297.6
h
t
240.1
290.1
340.I
390.I
440.1
490.1
54o.i
t
25
V
16.30
17 .60
18.86
20.10
21.32
22.55
2377
V
25
h
1160.4
1 184. 4
1208.2
12319
12556
1279.2
1302 .8
h
t
250.4
300.4
350.4
400.4
45o.4
500.4
55o.4
t
30
V
1374
1483
15.89
16.93
17.97
18.99
20.00
V
30
h
1163.9
1188.1
1212 .1
1236 .0
12597
1283.4
13071
h
t
2593
3093
3593
4093
4593
5093
5593
t
35
V
11.89
12.85
1375
14.65
1554
16.42
17.30
V
35
h
1166.8
1191.3
12154
1239.4
1263.3
1287. 1
1310.8
h
t
267.3
3173
367.2
4173
4673
5173
5673
t
40
V
10.49
n33
12.13
12.93
13.70
14.48
1525
V
40
h
1169.4
1 194.0
1218.4
1242 .4
1266.4
1290.3
1314.1
h
t
2745
3245
3745
4245
4745
5245
5745
t
45
V
939
10.14
10.86
n57
12 .27
12 .96
1365
V
45
h
1171 .6
1196.6
1221 .0
1245.2
1269.3
1293.2
1317.0
h
t
281 .0
33IO
381.0
431.0
481 .0
53i o
581.0
t
50
V
8.51
9.19
9.84
10.48
11 .11
11.74
12.36
V
5o
h
1173.6
1198.8
1223.4
1247.7
1271 .8
1295.8
I3I97.
h
t
287.1
3371
3871
4371
487.1
5371
587.1
t
55
V
7.78
8.40
9.00
959
10.16
10.73
11.30
V
55
h
H754
1 200 . 8
1225.6
1250.0
1274.2
1298. 1
1322.0
h
t
292 .7
342.7
392.7
442.7
492.7
542.7
592.7
t
60
V
7.17
775
8.30
8.84
936
9.89
10.41
V
60
h
1177.0
1202 .6
1227.6
1252. 1
1276.4
1300.4
13243
h
t
298.0
348.0
398 o
448.0
498.0
548.0
598.0
t
65
V
6.65
7.20
7.70
8.20
8.69
9.17
965
V
65
h
1178.5
1204.4
1229.5
1254.0 .
1278.4
1302.4
1326.4
h
t
302.9
352.9
402.9
452.9
502.9
552.9
602.9
t
70
v.
6.20
6.71
7.18
765
8. 11
8.56
9.01
V
70
h
1179.8
1205.9
1231 .2
12558
1280.2
13043
1328.3
h
t
307.6
3576
407.6
4576
507.6
5576
607.6
t
75
V
5.8i
6.28
6.73
7.17
7.60
8.02
8.44
V
75
h
1181.1
1207.5
1232 .8
12575
1282 .0
1306. 1
1330  1
h
t
312.0
362 .0
412 .0
462 .0
512.0
562.0
612 .0
t
80
V
547
592
634
675
7.17
7.56
795
V
80
h
1182.3
1208.8
1234.3
1259.0
1283.6
1307.8
I33L9
h
t
3i63
366.3
416.3
466.3
516.3
566.3
616.3
t
85
V
5i6
559
699
6.38
6.76
7.14
751
V
85
h
1183.4
1210. 2
12358
1260.6
1285.2
I3094
13335
h
t =
Temperature, deg. Fahr. v = Specific volume, in cubic feet, per pou
tid.
h = Tot£
d heat froi
n water at
32 degree
s, B.t.u.
APPENDIX
Table IV. — Continued
203
Pressure,
Satu
Degrees of Superheat.
Pressure,
Pounds
Absolute.
rated
Steam.
Pounds
Absolute.
50
IOO
150
200
250
300
t
320.3
370.3
420.3
470.3
520.3
570.3
620.3
t
90
V
4.89
529
567
6.04
6.40
6.76
7
11
V
90
h
1184.4
I2II .4
1237.2
1262 .0
1286.6
1310.8
1334
9
h
t
3241
3741
424.1
4741
5241
5741
624
1
t
95
V
465
503
539
574
6.09
643
6
76
V
95
h
1185.4
1212 .6
1238.4
1263.4
1288. 1
1312.3
1336
4
h
t
327.8
3778
427.8
4778
527.8
5778
627
8
t
IOO
V
443
479
5i4
547
5.80
6.12
6
44
V
IOO
h
1186.3
1213.8
1239.7
1264.7
1289.4
13136
1337
8
h
t
3314
381.4
4314
481.4
5314
581.4
631
4
t
105
V
423
458
4.91
523
554
58 5
6
15
V
105
h
1187.2
1214.9
1 240 . 8
1265.9
1290.6
13*49
1339
1
h
t
3348
3848
4348
484.8
5348
584.8
634
8
t
no
V
405
438
4.70
5oi
53i
56i
5
90
V
no
h
1188.0
1215.9
1242 .0
1267. 1
1291 .9
1316.2
1340
4
h
t
338.1
388.1
438.1
488.1
538.1
588.1
638
1
t
115
V
388
4.20
45i
4.81
509
538
5
66
V
II 5
h
1188.8
1216.9
1243. 1
1268.2
1293.0
I3I73
1341
5
h
t
3413
3913
4413
4913
5413
5913
641
3
t
120
V
373
4.04
433
4.62
4.89
5i7
5
44
V
120
h
1189.6
1217.9
1244. 1
1269.3
1 294. 1
1318.4
1342
7
h
t
' 3444
3944
444.4
494.4
5444
5944
644
4
t
125
V
358
3.88
4i7
445
47i
497
5
23
V
I2 5
h
1190.3
1218.8
1245. 1
1270.4
1295.2
I3I95
1343
8
h
t
3474
3974
4474
4974
5474
5974
647
4
t
130
V
345
374
4.02
4.28
454
4.80
5
05
V
130
h
1191 .0
1219.7
1246. 1
1271 .4
1296.2
1320.6
1344
9
h
t
35o.3
400.3
45o.3
500.3
55o.3
600.3
650
3
t
135
V
333
3.61
3.88
4.14
4.38
463
4
87
V
J 35
h
1191 .6
1220.6
1247.0
1272.3
1297 .2
1321 .6
1345
9
h
t
353  1
403 1
4531
503 1
5531
603.1
653
1
t
140
V
322
349
375
4.00
4.24
4.48
4
7i
V
140
h
1192 .2
1221 .4
1248.0
12733
1298.2
1322 .6
1346
9
h
t
3558
405.8
4558
5058
5558
605.8
655
8
t
145
V
3.12
338
3 63
387
4.10
433
4
56
V
J 45
h
1192 .8
1222 .2
1248.8
1274.2
1 299. 1
1323.6
1347
9
h
t
358.5
408.5
458.5
508.5
558.5"
608.5
658
5
t
150
V
3.01
327
35i
375
397
4.19
4
41
V
150
h
H934
1223 .0
1249.6
1275. 1
1300.0
13245
1348
8
h
t
361.0
411 .0
461 .0
511.
561.0
611 .0
661
t
155
V
2 .92
3*7
34i
3t>3
385
4.06
4
28
V
!55
h
1 194.0
1223 .6
1250.5
1276.0
1300.8
13253
1349
7
h
t
3636
413.6
4636
5136
5636
613.6
663
6
t
160
V
2.83
307
33o
353
374
395
4
15
V
160
h
"945
1224.5
1251.3
1276.8
1301.7
1326 . 2
1350
6
h
t
366.0
416.0
466.0
516.0
566.0
616.0
666
t
165
V
275
2.99
3.21
343
3 64
3 84
4
04
V
165
h
1195.0
1225 . 2
1252.0
1277.6
1302.5
13271
i35i
5
h
t
368.5
418.5
468.5
518.5
568.5
618.5
668
5
t
170
V
2.68
2 .91
3.12
334
354
373
3
92
V
170
h
II954
1225.9
1252.8
1278.4
I3033
1327.9
1352.3
h
t — Temperature, deg. Fahr. v = Specific volume, in cubic feet, per pound.
h = Total heat from water at 32 degrees, B.t.u.
204
APPENDIX
Table IV. — Continued
Pressure,
Satu
Degrees of Superheat.
Pressure.
Pounds
T~%
Absolute.
IdLLU
Steam.
irounas
Absolute
So
100
150
200
250
300
t
370.8
420.8
470.8
520.8
57o.8
620.8
670.8
t
175
V
2 .60
2.83
3 04
324
344
3 63
382
V
175
h
H959
1226.6
1253.6
1279. 1
13041
1328.7
13532
h
t
3731
423.1
4731
5231
5731
623.1
673.1
t
180
V
253
275
2 .96
3.16
335
354
372
V
180
h
1196.4
1227. 2
12543
1279.9
1304.8
13295
13539
h
t
3754
4254
4754
5254
5754
625.4
6754
t
185
V
2.47
2.68
2.89
3.08
327
345
3 63
V
185
h
1196.8
1227.9
12550
1280.6
13056
1330.2
13547
h
t
3776
427.6
477.6
527.6
5776
627.6
677.6
t
190
V
2 .41
2 .62
2.81
3.00
319
337
355
V
190
h
H973
1228.6
12557
1281.3
1306.3
1330.9
13555
h
t
3798
429.8
479.8
529.8
5798
629.8
679.8
t
195
V
235
255
2.75
293
3ii
329
3 46
V
i95
h
1197.7
1229.2
1256.4
1282 .0
1307.0
13316
1356.2
h
t
381.9
4319
481.9
5319
581.9
631.9
681.9
t
200
V
2 .29
2.49
2.68
2.86
304
3.21
338
V
200
h
1198.1
1229.8
1257. 1
1282 .6
1307.7
1332.4
1357
h
t
384.0
434 o
484.0
534
584.0
634.0
684.0
t
205
V
2.24
244
2 .62
2.80
2.97
3 14
3 30
V
205
h
1198.5
1230.4
12577
1283.3
1308.3
i333o
13577
h
t
386.0
436.0
486.0
536.o
586.0
636.0
686.0
t
210
V
2 .19
2.38
2.56
2.74
2 .91
307
323
V
210
h
1198.8
1 231 .0
1258.4
1284.0
1309.0
13337
1358.4
h
t
388.0
438.0
488.0
538.o
588.0
638.0
688.0
t
215
V
2 .14
233
2.51
2.68
2.84
3.00
3.16
V
215
h
1199.2
1231 .6
1259.0
1284.6
1309 . 7
13344
1359 1
h
t
3899
4399
489.9
5399
5899
6399
689.9
t
220
V
2 .09
2.28
245
2 .62
2.78
294
3.10
V
220
h
1199.6
1232 .2
1259.6
1285 .2
i3 IO 3
1335 1
13598
h
t
39 J 9
441.9
491.9
5419
5919
641.9
691.9
t
225
V
2.05
2.23
2 .40
257
2 .72
2.88
303
V
225
h
1199.9
1232.7
1260.2
1285.9
1310.9
13357
1360.3
h
t
3938
4438
4938
5438
5938
6438
6938
t
230
V
2 .00
2.18
235
. 2.51
2 .67
2.82
2.97
V
230
h
1200.2
1233.2
1260.7
1286.5
1311.6
1336.3
1361 .0
h
t
3956
4456
4956
5456
5956
645.6
695.6
t
235
V
1 .96
2 .14
2.30
2.46
2 .62
277
2 .91
V
235
h
1 200 . 6
12338
1261 .4
1287. 1
1312 . 2
I3370
1361.7
h
t
3974
4474
4974
5474
5974
647.4
697.4
t
240
V
1 .92
2 .09
2 .26
2 .42
257
2 .71
2.85
V
240
h
1200.9
12343
1261 .9
1287.6
1312.8
13376
1362.3
h
t
3993
4493
4993
5493
5993
6493
6993
t
245
V
1.89
2.05
2 .22
237
2.52
2.66
2.80
V
245
h
1 201 .2
1234.8
1262 .5
1288.2
I3I33
1338.2
1362.9
h
t
401 .0
451.0
501.0
55i o
601 .0
651.0
701 .0
t
250
V
185
2 .02
2 .17
233
247
2 .61
275
V
250
h
1201.5
12354
1263 .0
1288.8
13*39
1338.8
13635
h
t
402 .8
452.8
502.8
552.8
602.8
652.8
702.8
t
255
V
1. 81
1.98
2 .14
2.28
243
2.56
2 .70
V
255
h
1 201 .8
12359
1263 .6
1289.3
13145
13393
13641
h
t =
Temperature, deg. Fahr. v = Specific volume, in cubic feet, per pound.
h = Total heat from water at 32 degrees, B.t.u.
APPENDIX
205
Table V. — SATURATED VAPOR OF SULPHUR DIOXIDE
Reproduced by permission from Peabody's " Steam and Entropy Tables."
English Units.
fii
uA
<u
G
»>
g
4>
G
0>«i
a
Temperatui
Degrees Fa
renheit.
Pressure,
Pounds pe
Square Inc
•G .
"o'g
n cr ■
<ut3
■w
Q> U
lH O
HH ft
>
Heat Equiva
of Interna
Work.
Heat Equival
of Externa
Work.
P. o*
c
!3
>
'0
a>
Pi
t
P
A
H
L
P
Apu
9
5
40
314
29
166
195
182
13
—0.0632
23.O
— 20
5 90
— 21
169
190
176
14
0.0447
12.7
O
io.35
13
172
185
170
15
—0.0268
754
IO
i34i
 9
173
182
167
15
—0.0182
593
20
I7I5
 5
174
179
164
15
—0.0098
4.72
30
21 .66
— I
176
177
162
15
—0.0016
3.8i
40
27.06
3
177
174
158
16
. 0064
3.10
50
3345
7
178
171
155
16
0.0144
2.58
60
40.98
11
179
168
152
16
0.0221
2. 11
70
4975
15
181
166
ISO
16
0.0297
1.78
Table VI. — PROPERTIES OF CARBON DIOXIDE
Reproduced from Marks' " Mechanical Engineers' Handbook."
.fe8
Heat Content
G
UO G
u O
G
13 13
S
3
"o
nperatu
grecs F,
renheit
J3 coii
13 n
; Equiv
Extern
Work
*o
>
§•3
w G £;
<ij 3 ra
^ O 3
of
of
u
*c3
G P<
u c3
u a 1
53^
Liquid.
Vapor.
>
G—
3 °
(D
Pi
m
►h O
03
t
£
h '
H
z.
P
.A^M
s
20
221 .O
2475
IOO.50
12525
IO9.O
16.3
0.4173
O.0513
308.O
— 16.00
101 .00
117
00
IOI
3
157
0.2918
O.0325
10
362.5
11.36
100.89
112
35
96
9
154
0.2450
— O.0227
20
421 .6
— 6.40
100.43
I06
83
91
8
150
. 2060
— O.OI26
30
488.8
— 1 .04
9943
IOO
47
86
7
14.2
0.1724
— 0.002I
40
5645
436
98.25
93
89
80
4
135
0.1444
O.O087
50
650.0
10.76
96.30
85
54
73
3i
12.2
0.1205
O.O205
60
744.o
1785
9354
75
69
64
90
10.8
0.0986
O.0334
70
847.0
26.02
8935
63
33
54
03
93
0.0816
O.0483
Table VII. — NAPERIAN LOGARITHMS
Reproduced by permission from Goodenough's " Properties of Steam and Ammonia."
e =2.7182818 log e = 0.4342945 = M
1.0
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2.0
2.1
2.2
2.3
2.4
2.5
2.6
2.7
2.8
2.9
3.0
3.1
3.2
3.3
3.4
3.5
3.6
3.7
3.8
3.9
4.0
4.1
4.2
4.3
4.4
4.5
4.6
4.7
4.8
4.9
5.0
5.1
5.2
5.3
5.4
5.5
5.6
0. 0000
0.09531
0.1823
0.2624
0.3365
0. 4055
0. 4700
0.5306
0.5878
0.6418
0.6931
0.7419
0.7884
0.8329
0.8755
0.9163
0.9555
0.9933
1.0296
1.0647
1.0986
1.1314
1.1632
1.1939
1.2238
1.2528
1.2809
1.3083
1.3350
1.3610
1.3863
1.4110
1.4351
1.4586
1.4816
1.5041
1.5261
1.5476
1.5686
1.5892
1.6094
1.6292
1.6487
1.6677
1.6864
1.7047
1.7228
1.6114
1.6312
1.6506
1.6696
1.6882
1.7066
1.7246
0.00995
0. 1044
0. 1906
0. 2700
0.3436
0.4121
0.4762
0.5365
0.5933
0.6471
0.6981
0.7467
0.7930
0.8372
0.8796
0.9203
0.9594
0.9969
1.0332
0682
1.1019
1.1346
1.1663
1.1969
1.2267
1.2556
1.2837
1.3110
1.3376
1.3635
1.3888
1.4134
1.4375
1.4609
1.4839
1.5063
1.5282
1.5497
1.5707
1.5913
0.01980
0.1133
0. 1988
0.2776
0.3507
0.4187
0. 4824
0.5423
0.5988
0.6523
0.7031
0.7514
0.7975
0.8416
0.8838
0.9243
0.9632
1.0006
1.0367
1.0716
1.1053
1.1378 1.1410
1.1694 1.1725
1.2000 1.2030
0.02956
0.1222
0.2070
0. 2852
0.3577
0.4253
0. 4886
0.5481
0. 6043
0.6575
0. 7080
0.7561
0.8020
0.8459
0.8879
0.9282
0.9670
1.0043
1.0403
1.0750
1.1086 1.1119
0.03922
0.1310
0.2151
0.2927
0.3646
0.4318
0.4947
0.5539
0.6098
0.6627
0.7129
0.7608
0.8065
0. 8502
0. 8920
0.9322
0.9708
1.0080
1.0438
1.0784
1.2296
1.2585
1.2865
1.3137
1.3403
1.3661
1.3913
1.4159
1.4398
1.4633
1.4861
1.5085
1.5304
1.5518
1.5728
1.5933
1.6134
1.6332
1.6525
1.6715
1.6901
1.7884
1.7263
1.2326
1.2613
1.2892
1.3164
1.3429
1.3686
1.3938
1.4183
1.4422
1.4656
1.4884
1.5107
1.5326
1.5539
1.5748
1.5953
1.6154
1.6351
1.6544
1.6734
1.6919
1.7102
1.7281
1.1442
1.1756
1.2060
1.2355
1.2641
1.2920
1.3191
1.3455
1.3712
1.3962
1.4207
1.4446
1.4679
1.4907
1.5129
1.5347
1.5560
1.5769
1.5974
1.6174
1.6371
1.6563
1.6752
1.6938
1.7120
1.7299
206
0. 04879
0. 1398
0.2231
0.3001
0.3716
0.4382
0.5008
0.5596
0.6152
0. 6678
0.7178
0.7655
0.8109
0. 8544
0.8961
0.9361
0.9746
1.0116
1.0473
1.0818
1.1151
1.1474
1.1787
1.2090
1.2384
1.2669
1.2947
1.3218
1.3481
3737
1.3987
1.4231
1.4469
1.4702
1.4929
1.5151
1.5369
1.5581
1.5790
1.5994
1.6194
1.6390
1.6582
1.6771
1.6956
1.7138
1.7317
0.05827
0. 1484
0.2311
0.3075
0.3784
0.4447
0.5068
0. 5653
0. 6206
0.6729
0. 7227
0.7701
0.8154
0. 8587
0.9002
0.9400
0.9783
1.0152
1.0508
1.0852
1.1184
1.1506
1.1817
1.2119
1.2413
1.2698
1.2975
1.3244
1.3507
1.3762
0. 06766
0. 1570
0.2390
0.3148
0.3853
0.4511
0.5128
0.5710
0.6259
0. 6780
0.7275
0. 7747
0.8198
0.8629
0.9042
0.9439
0.9821
1.0188
1.0543
1.0886
1.1217
1.1537
1.1848
1.2149
1.2442
1.2726
1.3002
1.3271
1.3533
1.3788
8
1.4012 1.4036 1.4061
1.4255
1.4493
1.4725
1.4951
1.5173
1.5390
1.5602
1.5810
1.6014
1.6214
1.6409
1.6601
1.6790
1.6974
1.7156
1.7334
0. 07696
0. 1655
0.2469
0. 3221
0.3920
0.4574
0.5188
0. 5766
0.6313
0.6831
0.7324
0.7793
0.8242
0.8671
0.9083
0. 9478
0. 9858
0.08618
0. 1739
0.2546
0.3293
0.3988
0.4637
0.5247
0. 5822
0.6366
0.6881
0.7372
0. 7839
0.8286
0.8713
0.9123
0.9517
0.9895
1.0225 1.0260
1.0578 1.0613
1.0919 1.0953
1.1249 1.1282
1.1569
1.1878
1.2179
1.2470
1.2754
1.3029
1.3297
1.3558
1.3813
1.4279
1.4516
1.4748
1.4974
1.5195
1.5412
1.5623
1.5831
1.6034
1.6233
1.6429
1.6620
1.6808
1.6993
1.7174
1.7352
1.4303
1.4540
1.4770
1.4996
1.5217
1.5433
1.5644
1.5851
1.6054
1.6253
1.6448
1.6639
1.6827
1.7011
1.7192
1.7370
1.1600
1.1909
1.2208
1.2499
1.2782
1.3056
1.3324
1.3584
1.3838
1.4085
1.4327
1.4563
1.4793
1.5019
1.5239
1.5454
1.5665
1.5872
1.6074
1.6273
1.6467
1.6658
1.6845
1.7029
1.7210
1.7387
Table VII. — Continued. NAPERIAN LOGARITHMS
1
2
3
4
5
6
7
8
9
5.7
5.8
5.9
1.7405
1.7579
1.7750
1.7422
1.7596
1.7766
1.7440
1.7613
1.7783
1.7457
1.7630
1.7800
1.7475
1.7647
1.7817
1.7492
1.7664
1.7834
1.7509
1.7681
1.7851
1.7527
1.7699
1.7867
1.7544
1.7716
1.7884
1.7561
1.7733
1.7901
6.0
1.7918
1.7934
1.7951
1.7967
1.7984
1.8001
1.8017
1.8034
1.8050
1.8066
6.1
6.2
6.3
1.8083
1.8245
1.8405
1.8099
1.8262
1.8421
1.8116
1.8278
1.8437
1.8132
1.8294
1.8453
1.8148
1.8310
1.8469
1.8165
1.8326
1.8485
1.8181
1.8342
1.8500
1.8197
1.8358
1.8516
1.8213
1.8374
1.8532
1.8229
1.8390
1.8547
6.4
6.5
6.6
1.8563
1.8718
1.8871
1.8579
1.8733
1.8886
1.8594
1.8749
1.8901
1.8610
1.8764
1.8916
1.8625
1.8779
1.8931
1.8641
1.8795
1.8946
1.8656
1.8810
1.8961
1.8672
1.8825
1.8976
1.8687
1.8840
1.8991
1.8703
1.8856
1.9006
6.7
6.8
6.9
1.9021
1.9169
1.9315
1.9036
1.9184
1.9330
1.9051
1.9199
1.9344
1.9066
1.9213
1.9359
1.9081
1.9228
1.9373
1.9095
1.9242
1.9387
1.9110
1.9257
1.9402
1.9125
1.9272
1.9416
1.9140
1.9286
1.9430
1.9155
1.9301
1.9445
7.0
1.9459
1.9473
1.9488
1.9502
1.9516
1.9530
1.9544
1.9559
1.9573
1.9587
7.1
7.2
7.3
L9601
1.9741
1.9879
1.9615
1.9755
1.9892
1.9629
1.9769
1.9906
1.9643
1.9782
1.9920
1.9657
1.9796
1.9933
1.9671
1.9810
1.9947
1.9685
1.9824
1.9961
1.9699
1.9838
1.9974
1.9713
1.9851
1.9988
1.9727
1.9865
2.0001
7.4
7.5
7.6
2.0015
2.0149
2.0281
2.0028
2.0162
2.0295
2. 0042
2.0176
2.0308
2.0055
2.0189
2.0321
2.0069
2.0202
2. 0334
2. 0082
2.0215
2. 0347
2.0096
2.0229
2.0360
2.0109
2.0242
2.0373
2.0122
2.0255
2.0386
2.0136
2.0268
2.0399
7.7
7.8
7.9
2.0412
2.0541
2.0668
2. 0425
2.0554
2.0681
2.0438
2.0567
2.0694
2.0451
2.0580
2.0707
2. 0464
2.0592
2.0719
2. 0477
2.0605
2.0732
2. 0490
2.0618
2.0744
2.0503
2.0631
2.0757
2.0516
2.0643
2.0769
2.0528
2.0656
2. 0782
8.0
2.0794
2.0807
2.0819
2.0832
2.0844
2.0857
2.0869
2.0881
2.0894
2.0906
8.1
8.2
8.3
2.0919
2.1041
21163
2.0931
2.1054
2.1175
2.0943
2.1066
2.1187
2.0956
2.1078
2.1199
2.0968
2.1090
2.1211
2. 0980
2.1102
2. 1223
2.0992
2.1114
2.1235
2. 1005
2.1126
2. 1247
2.1017
2.1138
2.1258
2. 1029
2.1150
2. 1270
8.4
8.5
8.6
2. 1282
2.1401
2.1518
2. 1294
2.1412
2.1529
2.1306
2.1424
2. 1541
2.1318
2.1436
2.1552
2.1330
2. 1448
2.1564
2.1342
2.1459
2.1576
2.1353
2.1471
2.1587
2.1365
2. 1483
2. 1599
2. 1377
2.1494
2.1610
2.1389
2.1506
2. 1622
8.7
8.8
8.9
2.1633
2. 1748
2.1861
2. 1645
2.1759
2.1872
2.1656
2. 1770
2. 1883
2.1668
2.1782
2. 1894
2.1679
2.1793
2. 1905
2.1691
2.1804
2.1917
2. 1702
2.1815
2. 1928
2.1713
2.1827
2.1939
2.1725
2.1838
2.1950
2.1736
2.1849
2.1961
9.0
2.1972
2.1983
2.1994
2.2006
2.2017
2.2028
2.2039
2.2050
2.2061
2.2072
9.1
9.2
9.3
2.2083
2.2192
2.2300
2.2094
2.2203
2.2311
2.2105
2.2214
2.2322
2.2116
2.2225
2.2332
2.2127
2.2235
2.2343
2.2138
2. 2246
2.2354
2.2148
2.2257
2.2364
2.2159
2.2268
2.2375
2.2170
2.2279
2.2386
2.2181
2.2289
2.2396
9.4
9.5
9.6
2. 2407
2.2513
2.2618
2.2418
2.2523
2.2628
2. 2428
2.2534
2.2638
2. 2439
2.2544
2.2649
2. 2450
2. 2555
2.2659
2.2460
2.2565
2. 2670
2.2471
2.2576
2.2680
2.2481
2.2586
2.2690
2.2492
2.2597
2.2701
2. 2502
2.2607
2.2711
9.7
9.8
9.9
2.2721
2.2824
2. 2925
2.2732
2.2834
2.2935
2.2742
2. 2844
2.2946
2.2752
2.2854
2.2956
2.2762
2.2865
2.2966
2.2773
2.2875
2.2976
2.2783
2. 2885
2.2986
2.2793
2.2895
2.2996
2.2803
2.2905
2.3006
2.2814
2.2915
2.3016
10.0
2.3026
207
Table VIII. — LOGARITHMS
Reproduced by permission from Goodenough's " Properties of Steam and Ammonia."
Nat.
Nos.
Proportional Parts.
1
2
3
4
5
6
7
8
9
1 2
3
4
5
6
7
8 9
10
0000
0043
0086
0128
0170
0212
0253
0294
0334
0374
4 8
12
17 21
25
29
33 37
11
0414
0453
0492
0531
0569
0607
0645
0682
0719
0755
4 8
11
15
19
23
26
30 34
12
0792
0828
0864
0899
0934
0969
1004
1038
1072
1106
3 7
10
14
17
21
24
28 31
13
1139
1173
1206
1239
1271
1303
1335
1367
1399
1430
3 6
10
13
16
19
23
26 29
14
1461
1492
1523
1553
1584
1614
1644
1673
1703
1732
3 6
9
12
15
18
21
24 27
15
1761
1790
1818
1847
1875
1903
1931
1959
1987
2014
3 6
8
11
14
17
20
22 25
16
2041
2068
2095
2122
2148
2175
2201
2227
2253
2279
3 5
8
11
13
16
18
21 24
17
2304
2330
2355
2380
2405
2430
2455
2480
2504
2529
2 5
7
10
12
15
17
20 22
18
2553
2577
2601
%25
2648
2672
2695
2718
2742
2765
2 5
7
9
12
14
16
19 21
19
2788
2810
2833
2856
2878
2900
2923
2945
2967
2989
2 4
7
9
11
13
16
18 20
20
3010
3032
3054
3075
3096
3118
3139
3160
3181
3201
2 4
6
8
11
13
15
17 19
21
3222
3243
3263
3284
3304
3324
3345
3365
3385
3404
2 4
6
8
10
12
14
16 18
22 '
3424
3444
3464
3483
3502
3522
3541
3560
3579
3598
2 4
6
8
10
12
14
15 17
23
3617
3636
3655
3674
3692
3711
3729
3747
3766
3784
2 4
6
7
9
11
13
15 17
24
3802
3820
3838
3856
3874
3892
3909
3927
3945
3962
2 4
5
7
9
11
12
14 16
25
3979
3997
4014
4031
4048
4065
4082
4099
4116
4133
2 3
5
7
9
10
12
14 15
26
4150
4166 4183
4200
4216
4232
4249
4265
4281
4298
2 3
5
7
8
10
11
13 15
27
4314
4330
4346
4362
4378
4393
4409
4425
4440
4456
2 3
5
6
8
9
11
13 14
28
4472
4487
4502
4518
4533
4548
4564
4579
4594
4609
2 3
5
6
8
9
11
12 14
29
4624
4639
4654
4669
4683
4698
4713
4728
4742
4757
1 3
4
6
7
9
10
12 13
30
4771
4786
4800
4814
4829
4843
4857
4871
4886
4900
1 3
4
6
7
9
10
11 13
31
4914
4928
4942
4955
4969
4983
4997
5011
5024
5038
1 3
4
6
7
8
10
11 12
32
5051
5065
5079
5092
5105
5119
5132
5145
5159
5172
1 3
4
5
7
8
9
11 12
33
5185
5198
5211
5224
5237
5250
5263
5276
5289
5302
1 3
4
5
6
8
9
10 12
34
5315
5328
5340
5353
5366
5378
5391
5403
5416
5428
1 3
4
5
6
8
9
10 11
35
5441
5453
5465
5478
5490 5502
5514
5527
5539
5551
1 2
4
5
6
7
9
10 11
36
5563
5575
5587
5599
56115623
5635
5647
5658
5670
1 2
4
5
6
7
8
10 11
37
5682
5694
5705
5717
5729 5740
5752
5763
5775
5786
1 2
3
5
6
7
8
9 10
38
5798
5809
5821
5832
5843
5855
5866
5877
5888
5899
12
3
5
6
7
8
9 10
39
5911
5922
5933
5944
5955
5966
5977
5988
5999
6010
1 2
3
4
5
7
8
9 10
40
6021
6031
6042
6053
6064
6075
6085
6096
6107
6117
1 2
3
4
5
6
8
9 10
41
6128
6138
6149
6160
6170
6180
6191
6201
6212
6222
1 2
3
4
5
6
7
8 9
42
6232
6243
6253
6263
6274
6284
6294
6304
6314
6325
1 2
3
4
5
6
7
8 9
43
6335
6345
6355
6365
6375
6385
6395
6405
6415
6425
1 2
3
4
5
6
7
8 9
44
6435
6444
6454
6464
6474
6484
6493
6503
6513
6522
1 2
3
4
5
6
7
8 9
45
6532
6542
6551
6561
6571
6580
6590
6599
6609
6618
1 2
3
4
5
6
7
8 9
46
6628
6637
6646
6656
6665
6675
6684
6693
6702
6712
1 2
3
4
5
6
7
7 8
47
6721
6730
6739
6749
6758
6767
6776
6785
6794
6803
1 2
3
4
5
5
6
7 8
48
6812
6821
6830
6839
6848
6857
6866
6875
6884
6893
1 2
3
4
4
5
6
7 8
49
6902
6911
6920
6928
6937
6946
6955
6964
6972
6981
1 2
3
4
4
5
6
7 8
50
6990
6998
7007
7016
7024
7033
7042
7050
7059
7067
1 2
3
3
4
5
6
7 8
51
7076
7084
7093
7101
7110
7118
7126
7135
7143
7152
1 2
3
3
4
5
6
7 8
52
7160
7168
7177 7185
7193
7202
7210
7218
7226
7235
1 2
2
3
4
5
6
7 7
53
7243
7251
7259
7267
7275
7284
7292
7300
7308
7316
1 2
2
3
4
5
6
6 7
54
7324
7332
7340
7348
7356
7364
7372
7380
7388
7396
1 2
2
3
4
5
6
6 7
208
Table VIII. — Continued. LOGARITHMS
Proportional Parts.
Nat.
Nos.
1
2
3
4
5
6
7
8
9
12 3
4
5
6
7
8 9
55
7404
7412
7419
7427
7435
7443
7451
7459
7466
7474
1 2 2
3
4
5
5
6 7
56
7482
7490
7497
7505
7513
7520
7528
7536
7543
7551
12 2
3
4
5
5
6 7
57
7559
7566
7574
7582
7589
7597
7604
7612
7619
7627
1 2 2
3
4
5
5
6 7
58
7634
7642
7649
7657
7664
7672
7679
7686
7694
7701
1 1 2
3
4
4
5
6 7
59
7709
7716
7723
7731
7738
7745
7752
7760
7767
7774
1 1 2
3
4
4
5
6 7
60
7782
7789
7796
7803
7810
7818
7825
7832
7839
7846
1 1 2
3
4
4
5
6 6
61
7853
7860
7868
7875
7882
7889
7896
7903
7910
7917
1 1 2
3
4
4
5
6 6
62
7924
7931
7938
7945
7952
7959
7966
7973
7980
7987
1 1 2
3
3
4
5
6 6
63
7993
8000
8007
8014
8021
8028
8035
8041
8048
8055
1 1 2
3
3
4
5
5 6
64
8062
8069
8075
8082
8089
8096
8102
8109
8116
8122
1 1 2
3
3
4
5
5 6
65
8129
8136
8142
8149
8156
8162
8169
8176
8182
8189
1 1 2
3
3
4
5
5 6
66
8195
8202
8209
8215
8222
8228
8235
8241
8248
8254
1 1 2
3
3
4
5
5 6
67
8261
8267
8274
8280
8287
8293
8299
.8306
8312
8319
1 1 2
3
3
4
5
5 6
68
8325
8331
8338
8344
8351
8357
8363
837C
8376
8382
1 1 2
3
3
4
4
5 6
69
8388
8395
8401
8407
8414
8420
8426
8432
8439
8445
1 1 2
2
3
4
4
5 6
70
8451
8457
8463
8470
8476
8482
8488
8494
8500
8506
1 1 2
2
3
4
4
5 6
71
8513
8519
8525
8531
8537
8543
8549
8555
8561
8567
1 1 2
2
3
4
4
5 5
72
8573
8579
8585
8591
8597
8603
8609
8615
8621
8627
1 1 2
2
3
4
4
5 5
73
8633
8639
8645J8651
8657
8663
8669
8675
8681
8686
1 1 2
2
3
4
4
5 5
74
8692
8698
8704 8710
8716
8722
8727
873S
8739
8745
1 1 2
2
3
4
4
5 5
75
8751
8756
87628768
8774
8779
8785
8791
8797
8802
1 1 2
2
3
3
4
5 5
76
8808
8814
8820 8825
8831
8837 8842
8848
8854
8859
1 1 2
2
3
3
4
5 5
77
8865
887118876 8882
8887
8893 8899
8904
8910
8915
1 1 2
2
3
3
4
4 5
78
8921
8927
8932
8938
8943
8949
8954
8960
8965
8971
1 1 2
2
3
3
4
4 5
79
8976
8982
8987
8993
8998
9004
9009
9015
9020
9025
1 1 2
2
3
3
4
4 5
80
9031
9036
9042
9047
9053
9058
9063
9069
9074
9079
1 1 2
2
3
3
4
4 5
81
9085
9090
9096
9101
9106
9112
9117
9122
9128
9133
1 1 2
2
3
3
4
4 5
82
9138
9143
9149
9154
9159
9165
9170
9175
9180
9186
1 1 2
2
3
3
4
4 5
83
9191
9196
9201
9206
9212
9217
9222
9227
9232
9238
1 1 2
2
3
3
4
4 5
84
85
9243
9248
9253
9258
9263
9269
9274
9279
9284
9289
1 1 2
2
3
3
4
4 5
9294
9299
9304
9309
9315
9320
9325
9330
9335
9340
1 1 2
2
3
3
4
4 5
86
9345
9350
9355
9360
9365
9370
9375
9380
9385
9390
1 1 2
2
3
3
4
4 5
87
9395
9400
9405
9410
9415
9420
9425
9430
9435
9440
1 1
2
2
3
3
4 4
88
9445
9450
9455
9460
9465
9469
9474
9479
9484
9489
1 1
2
2
3
3
4 4
89
9494
9499
9504
9509
9513
9518
9523
9528
9533
9538
1 1
2
2
3
3
4 4
90
9542
9547
9552
9557
9562
9566
9571
9576
9581
9586
1 1
2
2
3
3
4 4
91
9590
9595
9600
9605
9609
9614
9619
9624
9628
9633
1 1
2
2
3
3
4 4
92
9638
9643
9647
9652
9657
9661
9666
9671
9675
9680
1 1
2
2
3
3
4 4
93
9685
9689
9694
9699
9703
9708
9713*
9717
9722
9727
1 1
2
2
3
3
4 4
94
9731
9736
9741
9745
9750
9754
9759
9763
9768
9773
1 1
2
2
3
3
4 4
95
9777
9782
9786
9791
9795
9800
9805
9809
9814
9818
1 1
2
2
3
3
4 4
96
9823
9827
9832
9836
9841
9845
9850
9854
9859
9863
1 1
2
2
3
3
4 4
97
9868
9872
9877
9881
9886
9890
9894
9899
9903
9908
1 1
2
2
3
3
4 4
98
9912
9917
9921
9926
9930
9934
9939
9943
9948
9952
1 1
2
2
3
3
4 4
99 9956
1
9961
9965
9969
9974
9978
9983
9987
9991
9996
1 1
2
2
3
3
3 4
209
2IO APPENDIX
Table IX
REFERENCES ON ENGINEERING THERMODYNAMICS
Engineering Thermodynamics. Charles E. Lucke. McGrawHill Book Co.,
New York City, N. Y.
Heat Engineering. Arthur M. Greene, Jr. McGrawHill Book Co., New York
City, N. Y.
Applied Thermodynamics. Wm. D. Ennis. D. Van Nostrand Co., New York
City, N. Y.
Principles of Thermodynamics. G. A. Goodenough. H. Holt & Co., New York
City, N. Y.
Thermodynamics of the Steam Engine and Other Heat Engines. Cecil H. Peabody.
John Wiley & Sons, Inc., New York City, N. Y.
Elements of Heat Power Engineering. Hirshfeld and Barnard. John Wiley &
Sons, Inc., New York City, N. Y.
Thermodynamics. Dr. Max Planck, Alex. Ogg, Trans. Longmans, Green & Co.,
New York City, N. Y.
Thermodynamics. De Volson Wood. John Wiley & Sons, Inc., New York City,
N. Y.
Technical Thermodynamics. Gustav Zeuner, J. F. Klein, Trans. D. Van Nos
trand Co., New York City, N. Y.
Thermodynamics of Heat Engines. Sidney A. Reeve. MacMillan Company,
New York City, N. Y.
An Introduction to General Thermodynamics. Henry A. Perkins. John Wiley &
Sons, Inc., New York City, N. Y.
Thermodynamics of Technical Gas Reactions. Dr. F. Haber. Longmans, Green
& Co., New York City, N. Y.
The Thermodynamic Principles of Engine Design. L. M. Hobbs. C. Griffin &
Co., London.
An Outline of the Theory of Thermodynamics. Edgar Buckingham. MacMillan
Company, New York City, N. Y.
The TemperatureEntropy Diagram. Charles W. Berry. John Wiley & Sons, Inc.,
New York City, N. Y.
The Steam Engine and Other Steam Motors. Robert C. H. Heck. D. Van
Nostrand Co., New York City, N. Y.
The Steam Engine and Turbine. Robert C. H. Heck. D. Van Nostrand Co.,
New York City, N. Y.
Steam Engine, Theory and Practice. Wm. Ripper. Longmans, Green & Co.,
New York City, N. Y.
Internal Combustion Engines. R. C. Carpenter, and Diederichs. D. Van Nos
trand Co., New York City, N. Y.
Gas, Petrol and Oil Engines, Vol. I. Dugald Clerk. John Wiley & Sons, Inc.,
New York City, N. Y.
Gas, Petrol and Oil Engines, Vol II. Clerk and Burls. John Wiley & Sons, Inc.,
New York City, N. Y.
The Design and Construction of Internal Combustion Engines. Hugo Guldner.
D. Van Nostrand Co., New York City, N. Y.
APPENDIX 211
Gas Engine Design. C. E. Lucke. D. Van Nostrand Company, New York City,
N. Y.
Modern Gas Engine and the Gas Producer. A. M. Levin. John Wiley & Sons,
Inc., New York City, N. Y.
The Gas Turbine. H. Holzwarth. C. G. Griffin Company, London.
Physical Significance of Entropy. J. F. Klein. D. Van Nostrand Co., New York
City, N. Y.
Practical Treatise on the "Otto" Cycle Gas Engine. Wm. Norris. Longmans,
Green & Co., New York City, N. Y.
Steam Tables and Diagrams. Marks & Davis. Longmans, Green & Co., New
York City, N. Y.
Properties of Steam and Ammonia. G. A. Goodenough, John Wiley & Sons, Inc.,
New York City, N. Y.
Tables of the Properties of Steam and Other Vapors and Temperature Entropy
Table. Cecil H. Peabody. John Wiley ^ Sons, Inc., New York City, N. Y.
Thermodynamic Properties of Ammonia. Robert B. Brownlee, Frederick G.
Keyes. John Wiley & Sons, Inc., New York City, N. Y.
Reflections on the Motive Power of Heat and on Machines fitted to Develop Power.
N. L. S. Carnot. Edited by R. H. Thurston, John Wiley & Sons, Inc., New
York City, N. Y.
Energy. Sidney A. Reeve. McGrawHill Book Co., New York City, N. Y.
A New Analysis of the Cylinder Performance of Reciprocating Engines. J. P.
Clayton, University of Illinois. Univ. of 111. Bulletin, Vol. IX, No. 26.
Urbana, 111. 191 2.
Thermal Properties of Steam. G. A. Goodenough. Bulletin of the University of
Illinois, Urbana, 111. Vol. XII, No. 1. 1914.
INDEX
Absolute temperature, 3, 15.
pressure, 6.
zero, 3.
Absorption system of refrigeration, 1
Adiabatic expansion, 27, 32, 36, 43,
no, 128.
compression, 32, 36, 43.
lines of steam, 109.
Air compressor, 178.
engines, 49.
Ammonia machine, 191.
Apu, 67.
Available energy of steam, 128134.
Barrel calorimeter, 84.
Barrus' calorimeter, 81, 82.
Boyle's law, 15, 29.
Brake horse power, 8.
Brayton cycle, 57.
British thermal unit, 3.
Calorie, 3.
Calorimeter, steam, 7785.
Carnot, 40.
Carnot cycle, 40, 117.
cycle, efficiency of, 44.
cycle, entropy changes, 94.
cycle, reversed, 45.
Centigrade, 2, 3.
Charles' law, 16.
Clausius, 119.
cycle, 119.
Clayton's analysis, 147.
Coefficient of flow, 165.
of performance, 194.
Combined diagrams, 140.
Combination law, 16.
Compound engine, 140.
Compressed air, 178185.
Compression, adiabatic, 32, 36, 43.
Compression, isothermal, 27, 42.
of gases, 26.
of vapors, 104.
93. Compressor, air, 178.
94, Condensing calorimeters, 84.
Conservation of energy, 9.
Conversion of temperatures, 2, 3.
Cycle, Brayton, 57.
Carnot, 40.
Clausius, 119.
definition of, 40.
hotair engine, 49.
internalcombustion engine, 53.
Otto, 56.
Rankine, 119, 140.
reversible, 45.
steam engine, 125.
Stirling, 50.
Density, 65, 154, 198.
Diagram, temperatureentropy, 95, 134.
heatentropy, in.
indicator, 26, 47, 140.
Mollier, 99, 113, (Appendix).
total heatentropy, 1 1 1 , 113.
Discharge, maximum, 135.
Dry saturated steam, 71.
Drying of steam by throttling, 75.
Efficiency, air engines, 52, 53.
Carnot cycle, 44.
Ericsson engine, 53.
Rankine cycle, 119, 140.
refrigerating machine, 94.
Stirling engine, 52.
thermal, n, 140.
volumetric, 181, 183.
Energy, available, 128134.
internal, 21, 35, 69.
intrinsic, 69.
213
214
INDEX
Engine, Ericsson, 52.
compound, 140.
hotair, 49.
Lenoir, 53.
Stirling, 50.
Entropy, 91.
diagram, 94102.
of the evaporation, 97.
of the liquid, 96.
Equivalent evaporation, 85.
Ericsson hotair engine, 52.
Evaporation, equivalent, 85.
external work of, 67.
factor of, 85.
internal energy of, 69.
latent heat of, 67.
Expansion of gases, 26.
of vapors, 104.
Expansions, adiabatic, 27, 32, 36, 43.
isothermal, 27, 28, 42, 94.
poly tropic, 112.
External work, 67.
work in steam formation, 67.
Factor of evaporation, 85.
First law of thermodynamics, 9.
Fliegner's formulas, 155.
Flow of air, 155.
in nozzles, 150.
in orifices, 150
of steam, 167, 171.
Footpound, 6.
Foot, square, 6.
Gas constant (it), 18, 22.
Gas, perfect, 14, 18.
Gramcalorie, 3.
Grashof, 168.
h (heat of liquid), 66.
H (total heat of steam), 68.
Heat engine efficiencies, 4060.
of liquid, 66.
units, 3.
Heatentropy diagram, 111.
Hirn's analysis, 144.
Hotair engine, 49.
Hyperbolic logarithms, 31, (Appendix).
Horse power, 8.
Indicator diagram, 26, 47, 140.
diagram, combined, 140.
Injectors, 165.
efficiency of, 166.
Internal energy, 21, 35, 69.
of evaporation, 69.
Intrinsic energy, 69.
Isothermal expansion, 27, 28, 42, 94.
compression, 27, 42.
lines of steam, 108.
Joule's law, 21. v
Kelvin, 21.
L (latent heat of steam), 67.
Latent heat of evaporation, 67.
Laws of perfect gases, 18.
of thermodynamics, 9.
Logarithms, natural, 31, (Appendix).
"common" or ordinary, 31.
Losses, turbine, 163.
Mean specific heat, 74, 75.
Mechanical equivalent of heat, 8.
Moisture in steam, 72, 77.
Mollier diagram, 99, 113 (Appendix).
Naperian logarithms, 31, (Appendix).
Napier's formula, 168.
Natural logarithms, 31, (Appendix).
Nonexpansive cycle, 127.
Nonreversible cycle, 45.
Nozzle, flow through, 150.
Nozzle losses, 173.
Nozzles, impulse, 162.
reaction, 164.
Orifice, flow through, 150.
Otto cycle, 56.
Per cent wet, 72.
Perfect gas, 14, 18.
INDEX
2I 5
Porous plug experiment, 21.
Power plant diagrams, 100.
Pressuretemperature relation, 16, 36.
units, 6.
Properties of gases, 197.
of steam, 63.
of vapors, 62.
g, 66.
Quality of steam, 72, no, 113.
R (thermodynamic or "gas" constant),
18, 22, 48.
r, 67.
Rankine cycle, 119, 140.
Ratio of expansion (r), 30.
of specific heats, 23, 154.
Receiver method, 158.
References, 198.
Refrigerating machines, 40, 178.
media, 87.
Refrigeration, applications of, 178.
unit of, 187.
Regenerator, 51.
Reversibility, 45.
Reversible cycle, 45.
Saturated steam, 62, 65.
Saturation curve, 142.
Scales, the rmo metric, 2, 3.
Second law of thermodynamics, 9.
Separating calorimeter, 82.
Small calorie, 3.
Specific heat, 4, 23.
heat at constant pressure, 5, 197.
heat at constant volume, 5, 197.
heat, instantaneous values of, 76.
heat, mean value of, 75.
heat, ratio of, 23.
heat of gases, 197.
heat of superheated steam, 75, 76.
heat of water, 4.
volume of gases, 6, 198.
volume of saturated steam, 65.
volume of superheated steam, 75.
Steam, dry, 71.
Steam engines, efficiencies of, 119.
engines, power plant, 100.
entropy for, 96.
saturated, 71.
superheated, 72, 73.
tables, (Appendix).
total heat of, 68.
turbine, 163.
wet, 71.
Stirling engine, 50.
Superheated steam, 72.
Superheating calorimeter, 77, 132.
Symbols, ix, x.
Tables, steam, (Appendix).
vapor, 63.
Temperature, absolute, 3.
entropy diagrams, 95102, 134.
entropy diagram for power plant,
100.
Thermal efficiency, n, 140.
unit, British, 3.
Thermodynamics, definition of, 1.
Thermo metric scales, 2, 3.
Throttling, 75.
Throttling calorimeter, 77, 80, 81.
Total heatentropy diagram, in, 113.
heat of saturated steam, 68.
heat of superheated steam, 73.
internal energy of steam, 69.
Turbine, steam, 163.
Underexpansion, 173.
Unit of heat, 3.
Vaporization (see evaporation).
Vapors, 62, 104.
properties of, 62.
Velocity, 155.
Volume, specific, 6, 65, 75, 198.
Volumetric efficiency, 181, 183.
Water, specific heat of, 4.
Watt, 8.
Weight, units of, 3 (footnote).
Wet steam, 71, no, 131.
2l6
INDEX
Wetness of steam, 72.
Wiredrawing, 75.
Work, external, 6, 67.
of adiabatic expansion, 67.
of Carnot cycle, 44.
of Clausius cycle, 120.
Work, of compression, 179.
of Rankine cycle, 120.
x (quality of steam), 72, no.
Zero, absolute, 3.
TOTAL HEATENTROPY DIAGRAM.
The ordinates are Total Heats.
The abscissae are Entropies.
Vertical lines are lines of constant entropy.
Horizontal lines are lines of constant total heat.
Reproduced by permission from
Marks and Davis' Steam Tables.
S
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